ultivating and Raising Families of Triangles dam arr, Julia Fisher, ndrew Roberts, David Xu, and advisor Stephen Kennedy March 13, 007
1 ackground and Motivation round 500-600.., either Thales of Miletus or Pythagoras of Samos introduced the Western world to the forerunner of Western geometry. fter a good deal of work had been devoted to the field, Euclid compiled and wrote The Elements circa 300.. In this work, he gathered a fairly complete backbone of what we know today as Euclidean geometry. For most of the,500 years since, mathematicians have used the most basic tools to do geometry a compass and straightedge. Point by point, line by line, drawing by drawing, mathematicians have hunted for visual and intuitive evidence in hopes of discovering new theorems; they did it all by hand. Judging from the complexity and depth from the geometric results we see today, it s safe to say that geometers were certainly not suffering from a lack of technology. In today s technologically advanced world, a strenuous effort is not required to transfer the capabilities of a standard compass and straightedge to user-friendly software. One powerful example of this is Geometer s Sketchpad. This program has allowed mathematicians to take an experimental approach to doing geometry. With the ability to create constructions quickly and cleanly, one is able to see a result first and work towards developing a proof for it afterwards. lthough one cannot claim proof by empirical evidence, the task of finding interesting things to prove became a lot easier with the help of this insightful and flexible visual tool. One of the most incredible features of Geometer s Sketchpad is its ability to produce dynamic constructions. In other words, one can create a construction with interdependent elements, alter one of the elements, and observe the resulting movement in the rest of the construction. These motions could not be easily represented otherwise (without the use of a flip book and a huge number of painstakingly accurate drawings). ecause of this, Geometer s Sketchpad has enabled modern mathematicians to ask questions that their predecessors were not able to formulate. On paper, geometry is static. With Geometer s Sketchpad, it becomes dynamic. With this tool in hand, let us begin our project with a few questions: Is there a way to naturally relate triangles? Is there a logical method to define families using intrinsic properties of triangles? an we easily construct these families? What would the triangle space that results from this construction look like topologically? How could we characterize motions through this space? In short, our task is to find a natural and meaningful way to name and relate triangles. Recognizing there are many ways to approach this task, we need a starting point. Since we want to find a way to characterize families of triangles that utilizes intrinsic triangle properties, it seems logical to begin with triangle centers. Thus, let us start with a brief overview. asic Triangle oncepts First, we should define a few very well-known triangle centers. lthough we are not including any proofs of their existence in our paper, these are simple to reproduce and can be found in any basic geometry text. Definition 1. median is straight line joining a triangle vertex to the midpoint of the opposite side. Proposition 1. The three medians intersect at a point. This point is known as the centroid and will be denoted by G. Definition. perpendicular bisector is a straight line that bisects a line segment at a right angle. Definition 3. The circumcircle is the circle determined by a triangle s three vertices. Definition 4. The circumradius is the radius of the circumcircle. 1
Definition 5. The circumcenter is the center of the circumcircle, denoted by O henceforth. Proposition. The three perpendicular bisectors of a triangle meet at the circumcenter. Definition 6. n altitude is a straight line going through a triangle vertex, perpendicular to the opposite side or extension of the opposite side. Proposition 3. The three altitudes of a triangle intersect at a point. This point is known as the orthocenter and will be denoted by H. In addition to the few presented here, there are a huge number of known triangle centers, many of which are constructed through quite elaborate processes. Interestingly, many of these triangle centers are collinear, lying on a line known as the Euler line. few of these include less common centers such as the nine-point center, the de Longchamps point, the Schiffler point, the Exeter point, and the far-out point. Most importantly for our purposes, the centroid, circumcenter, and orthocenter all lie on the Euler line. dditionally, it is a well-known fact that G always lies 1 3 of the way from O to H, a relationship of which we will frequently make use. ecause of its connection to triangle centers (and the fact that pril 15 th 007 is Euler s 300th birthday), it seems natural to use the Euler line as the basis of our construction. Such a construction would take full advantage of much work already done, since it is exploits the work already done on triangle centers. 3 lassifying Triangles ased on the Euler line Now, we can begin our search for a method of constructing a triangle based on a given Euler line. In the end, we want to develop a general method to construct all possible triangles. So, we begin by classifying all triangles that share the same Euler line. onsider the following question: Given the circumcenter, the centroid, and one vertex of a triangle, can you construct a triangle? Is the triangle with these three points unique? re there other triangles with different vertices but the same circumcenter and centroid? In order to address the first question, we need the known result below. Our own proof of it is provided. M c M b G M a Figure 1: n illustration of Proposition 4. ' Proposition 4. Given and its centroid G, let M a be the midpoint of segment, M b the midpoint of segment, and M c the midpoint of segment. Then, G = GM a, G = GM b, and G = GM c.
We will prove this statement for segment G. The proofs for segments G and G are identical. Proof. Extend segment M a past M a and mark point such that M a = M a. Since by construction M a = Ma and by the Vertical ngle Theorem M a = M a, it follows from SS ongruence that M a = M a. Thus, M a = Ma. y Proposition 7 of Euclid s Elements, this implies that. So, M b G = G. gain by the Vertical ngle Theorem, GM b = G. Thus, G M b G by Similarity. Since G = = M b, this and the previous triangle similarity imply that = = =. GM b M b M b Therefore, G = GM b. With this information in hand, we will show that with a given circumcenter, centroid, and one vertex of a triangle, we can construct a triangle. 3.1 Triangle onstruction Let line l and a point O on l be given. hoose a point not on l and create circle c centered at O with radius O. O will be the circumcenter of our triangle and one of its vertices. Thus, the other two vertices of our triangle will also lie on circle c. Let M and N be the intersections of l with c. Now, choose a point G O on l such that G lies between M and N. G will be the centroid of our triangle. onstruct point H on l such that OG = 1 3OH. y construction, H must be the orthocenter of the triangle. Now, construct the line that passes through and G and the line that passes through and H. Since it was just proven that the distance from one vertex of a triangle to its centroid is twice the distance from the centroid to the midpoint of the opposite side, construct point M a on G such that GM a = 1 G and such that M a is not between and G. Then, construct line m through M a such that m H. Let and be the intersections of m with circle c. Since H is the orthocenter of the triangle, H must be the altitude from vertex. Moreover, we constructed line m to be perpendicular to H. Thus, line m must be the side of the triangle opposite vertex, and and will be the triangle s remaining two vertices. Therefore, we have constructed a from a given Euler line l, two points on l, and a point not on l. c M l O G H N m M a Figure : The elements of the triangle construction. Removing the condition that the specific point must be a vertex of the triangle produces an infinite number of triangles that have circumcenter O and orthocenter H. So, we will choose to classify triangles according to their circumcenter, centroid, and a given circumradius. s a convention, we will orient ourselves with the Euler line as the traditional x-axis unless otherwise stated. 3
Definition 7. Let with circumcenter O and centroid G be given, and let T be the space of all triangles. Then, define a mapping F : T R 3, where F ( ) = (θ, g, r) with θ, g, and r determined as follows: θ = m GOP, where vertex P lies on the one-vertex side, g = OG, r = O. r O θ g G H Figure 3: OG = g, O = r, and GOP = θ where P is the vertex on the one-vertex side; in the diagram above, P = Note that in the event that one vertex falls on the Euler line we will have two one-vertex sides, and in that case θ will be the smaller of the angles. 3. Special ases of the onstruction The natural question to ask now is, is this construction well-defined? efore we address this, we must first discuss some special cases of our construction. These will help us prove that the construction is in fact well-defined. M b G O M c M a Figure 4: When g = 0, O and G are coincident. Proposition 5. is equilateral if and only if g = 0. 4
Proof. ( ) Let be equilateral. Let M a and M b be the midpoints of sides and, respectively. Then, construct lines M a and M b. Since =, M a = Ma, and =, it follows from SS congruence that M a = Ma. Therefore, m M a = m M a = π. Thus, M a is the perpendicular bisector of. onsequently, the circumcenter O of lies on M a. Moreover, by construction, the centroid G of lies on M a. y the same argument, O and G also lie on M b. Since M a and M b only intersect once, the only way that this can occur is if O = G. Thus, OG = g = 0. ( ) Let be given with OG = 0. Let M a and M b be defined as above. Now, construct G. Since G is the centroid, M a must lie on G. Since O = G, M a = M a. Thus, since M a is common, by SS congruence, M a = M a. This implies that =. y the same argument, =. Therefore, = =, and is equilateral. Note that an equilateral triangle does not in fact have an Euler line, since the triangle centers determining the Euler line collapse a single point. lso, since G and O are the same point the angle formed by O, G, and a vertex does not technically exist. Thus, we will define θ = π for reasons that will become clear later. Proposition 6. Let with orthocenter H be given. If H lies on the circumcircle of, then H is coincident with one vertex of. Proof. ( ) Let circle c centered at O be given, and let H and be points on c such that H. We will construct with circumcenter O, circumcircle c, and orthocenter H and show that H must coincide with either or. To begin, connect to the centroid G, which lies on the Euler Line one third of the way from O to H. Extend G beyond G one-half the length of G, and call the endpoint of this extended segment M a. y Proposition 4, M a must be the midpoint of. Since H is the orthocenter of, H must be perpendicular to. Thus, if we construct the line M a P, where P is the point on H such that M a P H, this line will contain vertices and. We will show that P = H. c O G H c H P M a Figure 5: P lies past H on ray H. To do this, suppose to the contrary that P H. Then, P lies between and H, past H on the ray H, or past on the ray H. ase 1: (Refer to Figure 5 above.) Suppose P lies past H on H. Then, vertex, the intersection of the circumcircle and the line M a P lies on the opposite side of the Euler line from 5
. y construction, m P = π. Now, construct H. H is the altitude from vertex and thus meets perpendicularly at H c. onsider H c. y angle subtraction, m H + m = π. onsider P. gain by angle subtraction, m P + m P = π. However, π = m P + m P > m H + m = π. This is a contradiction. Thus, P cannot lie past H on H. c P O G H H c M a Figure 6: P lies between and H. ase : (Refer to Figure 6 above.) Suppose P lies between and H. Then, vertex, the intersection of the circumcircle and the line M a P lies on the same side of the Euler line as. s in ase 1, since H is the altitude from, m P = π. Thus, m P +m P = π. onstruct H. Since H is the altitude from to, it meets perpendicularly at H c. It follows that m H c + m H c = π. However, π = m P + m P < m H c + m H c = π. This is a contradiction. Thus, P cannot lie between and H. P c H c O G H M a Figure 7: P lies past on ray H. ase 3: (Refer to Figure 7 above.) Suppose P lies past on the ray H. s in the previous two cases, m P = π. Thus, m P +m P = π. This implies that m = m P + π > π. onstruct H. Since H is the altitude from to, it meets perpendicularly at H c. Thus, m H c = m < π. This is a contradiction. Thus, P cannot lie past on ray H. Therefore, P is coincident with H. y the initial discussion, M a H contains the vertices and of. Since these vertices must lie on the circumcircle c, and H is on the circumcircle, it 6
follows that H must coincide with either or. orollary 1. Let with orthocenter H be given. Then, H is coincident with vertex if and only if m = π. Proof. ( ) ssume that H is coincident with vertex. It follows immediately from the work in Proposition 6 that m = π. ( ) ssume that m = π. Then, is the altitude from vertex to, and is the altitude from vertex to. Since the altitudes of a triangle coincide at H, it follows that and H are coincident. Proposition 7. Triangle is a right triangle if and only if g = r 3. Proof. ( ) If g = r 3, then OH = r. That is, H lies on the circumcircle. Therefore, by Proposition 6 and orollary 1, is a right triangle. ( ) If is a right triangle, then by orollary 1, H is coincident with one vertex. Without loss of generality, let that vertex be vertex. Thus, H is on the circumcirle, so OH = r. Since G is one-third of the way from O to H, it follows that g = r 3. O G H Figure 8: Right triangles occur when g = r 3 and H lies on the circumcircle. To follow Proposition 7, we have the next two propositions which deal with the cases in which g > r 3 and g < r 3. Proposition 8. Triangle is an obtuse triangle if and only if g > r 3. Proof. We will begin with the reverse direction. ( ) Let g > r 3. Then, since OH = 3OG = 3g > 3( r 3 ) = r, it follows that the orthocenter H lies outside the circumcircle. onstruct the altitude from vertex. Without loss of generality, assume that H lies past vertex on ray. Since m H = π, it follows that m H < π. Thus, m > π. Therefore, is obtuse. ( ) Let be obtuse. Without loss of generality, assume that m > π. onstruct the altitude from vertex. It will intersect side either past on ray or past on ray by the following: Suppose that the altitude from vertex intersects segment. Let H a be its intersection with. Then, m H a = π. y assumption, m > π. Thus, m H a + m + m H a > π. This is a contradiction. Thus, the altitude from vertex 7
intersects an extention of segment. y the same argument, the altitude from vertex intersects an extention of segment. Now, the extentions of and must lie outside the circumcircle of. Therefore, the orthocenter H, the intersection of the altitudes from vertices and, must lie outside the circumcircle. Since OH = 3OG = 3g, and since we now know that OH > r, it follows that 3g > r or g > r 3. Proposition 9. Triangle is an acute triangle if and only if g < r 3. Proof. This follows from a process of elimination using Propositions 7 and 8. The three propositions above are nice because they indicate that the construction cleanly separates acute, right, and obtuse triangles. We will next examine another type of triangles isosceles triangles. Proposition 10. If is isosceles, one of its vertices lies on the Euler line. Proof. ssume that is isosceles with =. Let M a be the midpoint of. Then, M a contains the centroid G of. Moreover, since =, M a = Ma, and M a is shared, it follows that M a = M a. Thus, M a = M a. Since they are supplementary, m M a = m M a = π. This implies that M a is the altitude of from point. Thus, the orthocenter H of must lie on M a. Therefore, M a coincides with the Euler line. ' '' O G G' G'' ' '' Figure 9: few different isosceles triangles. Proposition 11. If is not a right triangle and has a vertex lying on the Euler line, then it is isosceles. Proof. ssume that the Euler line l of intersects vertex. Since the centroid of lies on l, it follows that l must intersect at the midpoint M of. Thus, M a = Ma. lso, the orthocenter of lies on l and is not coincident with. Thus, m M a = m M a = π. Finally, by SS congruence, M a = M a (since side M a is shared). This implies that =. 8
3.3 Showing that the onstruction is Well-Defined We are finally ready to show that the construction is well-defined. Proposition 1. The mapping F : T R 3 is well-defined. D M b M e O M c O' M f G G' F M a M d E Figure 10: Showing the construction is well-defined. Proof. To show our construction is well-defined we must show that given two triangles and DEF, = DEF if and only if the coordinates (θ, g, r) of equal the coordinates (θ, g, r ) of DEF. The reverse direction is covered since given two sets of equal parameters, we know that the corresponding triangles will be congruent by construction. Thus,we only have to show that two congruent triangles will produce the same parameters. ssume = DEF with corresponding to D, corresponding to E, and corresponding to F. onstruct the centroids and circumcenters of the given triangles. Since the triangles are congruent it must be that corresponding vertices lie on the same side of the Euler line with respect to the other vertices. ase i) Without loss of generality let and be on the same side of the Euler line. Then, D and E must also lie on the same side of the Euler line. lso, it must be that r = r, since otherwise the triangles would have different circumcircles. Therefore, O = O F in the diagram. lso we know the corresponding medians must be congruent by simple SS arguments. Now, G = Mc 3 and G F = F M f 3. ut M c = F Mf because they are corresponding medians. Therefore, G = G F. y a similar argument, GM a = G M d. So, by SSS GM a = F G M d, which gives us GM a = G F M d. lso, by a hypotenuse-leg right-triangle congruence argument, OM a = O F M d. Therefore, OM a = O F M d. y subtraction, GO = G F O. So, by S congruence, GO = G O and GO = G O. Thus, it must be that g = g. It also follows that GO = G O F, so θ = θ. We now have to consider the cases in which we have a vertex on the Euler line. First, we will consider the right triangle case. ase ii) If we let m = π = m DEF, then r = = DF = r. Since we have right triangles, g = r 3 = g. lso O = O DE by SSS. So O = EO D. If O π, then θ = m O = θ. Otherwise, O > π, which means θ = π m O = θ. ase iii) If we have congruent isosceles triangles, the arguments for cases i) and ii) hold, except that it does not matter which vertex you choose to form θ as long as it is not the vertex on the Euler line. 9
ase iv) When we have congruent equilateral triangles, r must equal r, θ (likewise θ ) does not actually exist or have any affect in determining the triangle and g = 0 = g. 4 The Topology of Triangle Space s mentioned in the introduction, one of our goals is to figure out what triangle space looks like. One could argue that this can be done by considering triangles in the traditional sense: defining a triangle by the lengths of its sides. In either case, you are given three numbers, and those three numbers correspond to a point in R 3. This point corresponds to the triangle defined by the three numbers. The problem with considering triangles in the traditional sense is that every triangle is corresponds to six different points (one for each permutation of the three sides). It is not a trivial problem to define the space in such a way that every triangle is represented only once. y defining a triangle the way we have, we have ensured that every triangle is represented by exactly one ordered triplet. However, not every point in R 3 represents a triangle. It will be our next task to determine exactly which values of θ, g, and r give us legitimate triangles. To help us reach our goal, let us first examine similar triangles. Theorem 1. Two triangles defined by (θ, g, r) and defined by (θ, g, r ) are similar if and only if θ = θ, g = kg, and r = kr for k > 0. Proof. ( ) Let be similar to. Let G and G be the centroids of and O and O the circumcenters of and, respectively. lso, let M a and M a be the midpoints of segments and, respectively. Let M b and M b be the midpoints of and, respectively. Since the two triangles are similar, it follows that = M a M a = k. Therefore, M a M a. So, M a M a = k. Now, G and G must lie on M a and M a,respectively. Moreover, GM a = 1 3 M a and G M a = 1 3 M a. GM So, a G M a = M a M a. lso, = M b = k. So, M M b b M M b. So, b M = k. ut, b GM b = 1 3 M b and G M b = 1 3 M GM b. Thus, b G M = k. This implies that M b b M a G M b M a by SSS similarity. Thus, GM b M a = G M b M a. Now, m O M b G + m G M b M a + m M a M b = π m GM b M a = m G M b M a m M a M b = m M b m GM b M a = m M b m G M b M a = m M a M b m OM b G = m O M b G Thus, by SS similarity, OM b G = O M b G. This implies that OG O G = GM b G M b = k. So, OG = ko G. ut, O G = g and OG = g. Thus, g = kg. M Now, we know that b = M a M b M a = k. So, M b M a M b M a by SS similarity. Therefore, M b M a = M b M a, M a M b = M a M b, and MMa M M a. Thus, by subtraction, OM a M = O M a M b and OM b M a = O M b M a. This implies that OM b M a O M b M a. OM So, b O M = M bm a b M b M a = k. Thus, OM b O M b O. Therefore, O = k. ut, O = r and O = r. So we know r = kr. Now, construct lines OG, O G, and segments O and O. We have shown that the segments connecting the centroids to the vertices are proportional by k, and OG = ko G. Thus, OG O G by SSS similarity. This implies that GO = G O. So, since GO = θ and G O = θ, θ = θ. 10
' r/k b/k r b θ θ O g G O g/k G' ' a a/k ' Figure 11: ngle θ is the same in both triangles. ( ) Let defined by (θ, g, r) and defined by (θ, g, r ) be two triangles with the property that r = kr, g = kg, and θ = θ. First, let M a be the midpoint of. Now, we can see that OG O G by SS similarity. Thus, = k. So, GM a G G M a M a = 3 G 3 = G G G Then, G M = k and GM a a = kg M a. lso, GO = G O and thus OGM a = O G M a. Therefore, OGM a O G M a OM by SS similarity. So, a O M = k and OM a a = ko M a. lso, OM a G = O M ag. ecause O and O lie on the perpendicular bisectors to sides and respectively, we know, = k M a = (kr) OM a ( M a) = r (O M a) M a = (kr) (ko M a) M a = k (r (O M a) ) M a = k ( M a) M a = k M a lso, = M a = k M a = k. Thus, M a M a M a = M a and by SS similarity. by SS similarity. Thus, Let us now begin looking at the bounds on θ, g, and r. Proposition 13. If (θ, g, r) represents a triangle, then r > 0. Proof. The only acceptable values of r are r > 0 because r = 0 would mean that the circumcircle is only a point, which does not allow for any triangles. Moreover, r < 0 is absurd. lso, if r is bounded 11
above, then we restrict ourselves to triangles with side lengths less than r, so r can be any positive number. That being said, Theorem 1 implies that r is just a scaling variable. Therefore we can get a clear picture of what triangle space looks like by examining cross-sections of R 3 parallel to the θg -plane. Proposition 14. If (θ, g, r) represents a triangle, then 0 g < r cos[ 1 (θ + sin 1 ( 1 sin θ))] cos[ 1 (θ sin 1 ( 1 sin θ))] θ O G min G' max ' ' Figure 1: Note that vertices and are coincident. Proof. s motivation for this proof, imagine being given a triangle inscribed in a circle. Take the centroid G of the triangle, which lies on the Euler line, and start moving it away from the circumcenter toward the closest edge of the circle. Since G is the intersection of the lines connecting the vertices of the triangle with the midpoints of the opposite sides, it follows that as G approaches the edge of the circle, two vertices of the triangle (say and ) begin to come closer together. This is because G is approaching one side of the triangle (say ), and as it gets closer and closer to that side, the line connecting vertex and the midpoint of, M a moves closer and closer to the edge of the circumcircle. Thus, when G reaches side, and consequently disappears. It seems that this point is the limit for OG or g. Therefore, we first identify a way to calculate this limit. Let M a again be the midpoint of the side opposite vertex. We want to find the value of g so that M a lies on the the circumcircle. Now, consider OM a. Let be the vertex of inscribed in the circle and M a be the point on ray G such that G = GM a. Let y = GM a and g = OG. Then, since OM a and O are both radii of the circle, it follows that OM is isosceles. Thus, OM a = sin θ OMa. Let α = M a OG and GO = θ. Now, by the Law of Sines, y = sin OM a g in OG. lso, sin α y = sin OM a g = sin OM a g in M a OG. Therefore, sin θ y that sin α = sin θ. This in turn implies that α = sin 1 1 sin θ. Now, π (θ+α) γ = OG, then γ = π θ β = π θ π + θ + α = π θ + α = sin α y, which implies = β. If we define. Finally, again by the Law of sin π Sines, θ α = sin π θ + α. g = r sin π θ α x r sin π θ + α realizing that sin( π φ) = cos φ provides the desired result,. Replacing α with sin 1 1 sin θ and g < r cos[ 1 (θ + sin 1 ( 1 sin θ))] cos[ 1 (θ sin 1 ( 1. sin θ))] 1
Now that we have g bounded above, we consider the lower bound. learly we need to include g = 0 because we need to include equilateral triangles. We have always looked at our triangles with G on the right side of O. If we look at the triangles we get with G on the left side of O we realize they are the same as the ones with G on the right side. They can be mapped to each other by reflecting across the line perpendicular to the Euler line at O. Thus, 0 g < r cos[ 1 (θ + sin 1 ( 1 sin θ))] cos[ 1 (θ sin 1 ( 1. sin θ))] The bounds for θ are a lot harder to sort out because they come in cases which depend on g For the following propositions, we will denote the θ value for an isosceles triangle with a vertex and G lying on the Euler line on the same side of O as θ R. lso, we will denote the θ value for an isosceles triangle with a vertex and G lying on the Euler line on the opposite sides of O as θ L. Proposition 15. If g < r 3, then cos 1 ( 3g+r r ) θ cos 1 ( 3g r r ). ' θ max θ min ' O G H ' Figure 13: The bounds are created by an isoceles triangle on either end. Proof. Since g < r 3, every triangle with a vertex on the Euler line is an isosceles triangle. onsider the isosceles triangle with vertex on the Euler line on the same side of O that G is on. So, θ R = m GO. Let M b be the midpoint of. Since this is an isosceles triangle, M b lies on the Euler line and m M b O = π. lso, r g = G = (GM b). Therefore, OM = r 3g r. We also know m OM = π θ R. So cos(π θ R ) = cos θ R = r 3g r. Therefore, θ R = cos 1 ( 3g r ). r consequence of this isosceles triangle case is that GO = O. lso, GO = GO, and both measure θ, so we can think of GO as θ. y the law of cosines, we know that O = cos 1 ( r () r ). 13
In the appendix, we prove that = 3r 3rg cos θ 3rg sin θ 3r 9g + 6rg cos θ r + 9g 6rg cos θ. When θ = cos 1 ( 3g r r ), (after some simplification) we see (θ m O) θ = 1 3g r. So, since g < r 3, (θ m O) > 0. θ This means that as we increase m GO, the difference m GO m O increases. So, m GO m O > 0. Therefore, and must be on the same side of the Euler line. This means that θ is made by GO, and GO < cos 1 ( 3g r ), r otherwise and would be on the same side of the Euler line (and cannot be on both sides of the Euler line). This proves that θ R is the upper bound for θ when g < r 3. Likewise, the lower bound for θ is found in the other isosceles triangle case. That is, when we have an isosceles triangle with a vertex (without loss of generality, ) on the opposite side of O from G. Let θ L be made by GO. gain, let M c be the midpoint of. Since this is an isosceles triangle, M c lies on the Euler line and m M c O = π. lso, g + r = G = (GM c). Therefore, OM c = 3g+r r. We also know m OG = θ L. So, θ L = cos 1 ( 3g + r ). r The consequence of this isosceles triangle case is that GO + O = π. lso, GO = GO, and both measure θ, so we can think of GO as θ. y the law of cosines, we know that In the appendix, we prove that = O = cos 1 ( r () r. 3r 3rg cos θ + 3rg sin θ 3r 9g + 6rg cos θ r + 9g 6rg cos θ. When θ = cos 1 ( 3g+r r ), (after some simplification) we see (θ + m O π) = (1 + 3g θ r ). So, (θ + m O π) < 0. θ This means that as we decrease m GO, the m GO + m O π decreases. So, m GO + m O < π. 14
Therefore, and must be on the same side of the Euler line. This means that θ is made by GO, and GO > cos 1 ( 3g + r ), r otherwise and would be on the same side of the Euler line (and cannot be on both sides of the Euler line). This proves that θ L is the lower bound for θ when g < r 3. ombining the upper and lower bounds we get, cos 1 ( 3g + r r ) θ cos 1 ( 3g r ). r Note that if we consider the g = 0 case, we have defined θ = π which falls between the bounds. Proposition 16. If g = r 3, then 0 < θ π. Proof. To find the upper bound for θ when g = r 3 we again look at the isosceles triangle with vertex on the same side of O as G. We know that in this case, O and M b are the same point, and that m OG = π. We also know that, O, and are collinear, so if we increase OG we necessarily decrease GO, and the two angles must sum to π. The right triangle case is the reason we had to add the requirement that θ be the smaller of the two angles if we have two one-vertex sides. In this case, the largest value that the smaller angle can take is π. When we go to look at the isosceles triangle with vertex on the opposite side of O from G, we realize that we cannot make this triangle because two vertices would have to fall on on the Euler line, which is impossible. So if we keep at the right angle, we can move vertex arbitrarily close to the Euler line on the opposite side of the circle from. This means that we can make m GO arbitrarily close to π, so m GO can be made arbitrarily close to 0. So when g = r 3 0 < θ π. Proposition 17. If g > r 3, then cos 1 ( 3g r r ) θ cos 1 ( 3g r rg ). Proof. gain we consider the isosceles triangle with vertex on the Euler line on the same side of O that G is on. So, m GO = θ R = cos 1 ( 3g r ). r s before we have GO = O. lso, GO = GO, and both measure θ, so we can think of GO as θ. Following the same procedure for the isosceles triangle with θ = θ R (from the g < r 3 case), we see, (θ m O) = 1 3g θ r. ut this time, since g > r 3, (θ m O) < 0. θ This means that as we increase m GO, the difference m GO m O decreases! So, m GO m O < 0. 15
' θ max θ min O G ' H, Figure 14: The bounds are created by an isoceles triangle and a line. ' Therefore, and must be on opposite sides of the Euler line. This proves that θ R bound for θ when g > r 3. is the lower When g > r 3 there is no other isosceles triangle to look at. This is because M b = 3 (r + g) > 3 ( 4r 3 ) = r, so the midpoint of the side opposite and therefore the side itself is not contained in the circumcircle. Even though this means that our usual means of finding the other bound is not possible, it gives us another possibility. Now we set out to find where the triangle disappears. To do this, we find the angle that forces M a to land on the circumcircle (also forcing, M, and G to be collinear. Therefore, we are interested in the case where O = r = OM a. This means OM a is an isosceles triangle, and OM a = Ma O. If we define l = G, then M a = 3l. y the law of cosines, r = r + 9l 4 3rl cos M ao. fter some simple algebra we discover, r(cos MO) = 3l. y the Law of osines in OG we know g = r + l rl cos OM a. Substituting and solving for l, we get l = (r g ). We again use the law of cosines in OG and get l = r + g rg cos θ. Substituting for l and solving for θ we get θ = cos 1 ( 3g r ). rg Thus, cos 1 ( 3g r r ) θ cos 1 ( 3g r ). rg The nice part about the last upper bound for θ is that it is the inverse function to our upper bound for g. With that in mind, here is a cross-section of our space for r = 1 : s you see in the picture, we have built in a flipping effect that happens when g crosses the r 3 line. This stems directly from our definition of θ as the angle formed to the one-vertex side. It is a problem we can live with and will deal with shortly, but we do so with the belief that our definition of θ is the 16
Figure 15: cross-section of the space when r = 1. most natural way to eliminate overcounting. First, we must quickly return to the situation where we fix θ and allow g to vary. If we do this and allow g to cross the r 3 threshold, it is not immediately clear what happens. Upon reflection of the g = r 3 case, we realize that (θ, g, r) = (π θ, g, r). Now that we know where all of our triangles are and what triangle space looks like, it is natural to define a metric to determine how close two triangles are to being congruent. We would like the metric to have two additional properties: i) ccount for the fact that all triangles with g = 0 and the same r are congruent, and ii) ccount for the flipping effect when g crosses r 3. First, we will account for the flipping by defining a new function, θ = { θ i π θ i r g i i 3 g i > r i. 3 Let s and t be triangles such that s = (θ 1, g 1, r 1 ) and t = (θ, g, r ). Define D(s, t) = (g 1 θ1 g θ ) + (g 1 g ) + (r 1 r ). ttaching the g s onto the θ terms accounts for how alike triangles with small g values are. It also leads to two nice propositions dealing with similar triangles, but first we will show a result about triangles who have the same r and g values. Definition 8. gr -family of triangles is a family of triangles with the same values of g and r. Proposition 18. If triangles s and t are in the same gr -family, with s = (θ 1, g, r) and t = (θ, g, r), then D(s, t) = g θ 1 θ. Proof. D(s, t) = (g θ 1 g θ ) + (g g) + (r r) = g ( θ 1 θ ) = g θ 1 θ. 17
Proposition 19. Let s and t be triangles such that s = (θ, g, r) and t = (θ, kg, kr). Then D(s, t) = k 1 (g θ) + g + r Proof. egin with the definition of D(s, t) = (kg θ g θ) + (kg g) + (kr r). Then factor a (k 1) from each term, and bring it outside the square root as a k 1. Proposition 0. Given triangles p = (θ 1, g, r), p = (θ 1, kg, kr), q = (θ, g, r), q = (θ, kg, kr), then D(p, q ) = kd(p, q). Proof. D(p, q ) = (kg θ 1 kg θ ) + (kg kg) + (kr kr) = kg θ 1 θ = kd(p, q). 5 Tracing Theorems Now that the foundation of our construction has been laid out, we turn our attention to the behavior of triangle families. Several interesting properties were found with great assistance from the animation feature of Geometer s Sketchpad, which allows us to study the behavior of two families of triangles. We first study the family of triangles obtained through varying g, which we call the rθ -family. ' M a M' a O G G' ' Figure 16: rθ -family: the parallel line. Proposition 1. Let be given with circumcenter O and centroid G. If we fix r, θ and move G along the Euler line, we get a family of triangles denoted as the rθ -family. Then the locus of midpoints M a for side forms a line segment parallel to the Euler line. 18
Proof. For a determined by θ, g and r in our triangle construction, let be the triangle formed by moving G away from O along the Euler ine. is determined by θ, g and r. Let M a be the midpoint of and G be the centroid of. Then, consider M a M a. y construction, G lies on M a, and G lies on M a. Now, we know that G = GM a and G = G M. Thus, with M a M a in common, by SS similarity, M a M a GG. This implies that GG = M a M a. Thus, GG M a M a. Therefore, as G moves away from O on the Euler line, M a moves along a line parallel to OG. O G M b M'' M c M p Figure 17: rθ -family: the circle. Proposition. In our rθ -family of triangles, the locus of midpoints M b for side and M c for side forms part of a circle centered at the midpoint M of O. Proof. Let O be the circumcenter of and let M b and M c be the midpoints of sides and respectively. lso, let M p be the midpoint of segment M c. Now, construct O and call its midpoint M. y SS similarity, OM c M M P. Thus, M M P is parallel to OM, which is perpendicular to since M is the midpoint of and O is the circumcenter. Now, we know that M M P is the perpendicular bisector of the segment M. From this, we know that M = M M. y the same reasoning, we know that M = M M. So, M is the circumcenter of M M. Definition 9. For any, the circle determined by the midpoints of each side also passes through the feet of its altitudes and the midpoints of the segments joining each vertex with the orthocenter. This circle is known as the nine-point circle, or the Euler circle. Proposition 3. Recall that the collection of triangles sharing a circumcircle O and centroid G is defined as the gr -family. This family of triangles shares the same nine-point circle, which is formed by the locus of the midpoints of their sides. Proof. onsider with M a, M b and M c as the midpoints of,, and respectively. Extend OG to O such that GO = OG. ecause GO = OG, GM = G and O GM = OG, O GM OG. Therefore, O M = O = r. y similar logic, O is equidistant to M a, M b and M c. Thus, as θ changes, M a, M b, and M c trace the nine-point circle. nother way to look at this is that since the gr -family of triangles share the same circumcircle and centroid, they should also share the collection of complementary points for the circumcircle, i.e., the nine-point circle. 19
M c M b O G O' H M a 6 The Symmedian Point Figure 18: gr -family: the nine-point circle. Having examined the loci of certain well-known points in our construction, we now turn our attention to a more obscure triangle center the symmedian point. In the collection of triangle centers, few could be considered well-known to geometers and even fewer to the average geometry student. Occasionally, however, centers which should be a part of mathematicians base knowledge disappear from the contemporary consciousness. The symmedian point is one such center. Well explored many years ago, the symmedian point has a plethora of useful and fascinating properties. In his work 19th and 0th entury Euclidean Geometry, Ross Honsberger calls it one of the jewels of modern geometry [Honsberger, pg. 53]. In order to begin our brief study of this geometric gem, we first need to understand some established definitions and theorems. Thus, let us define the concept of isogonal conjugates. Definition 10. Let be given. Then, the isogonal conjugate of P, where P is any point in the plane, is the reflection of P over the angle bisector of. P and its reflection are called isogonal conjugate lines or simply isogonal conjugates. Figure 19: Isogonal conjugate lines. In Figure 19 above, the middle line is the angle bisector of, and the two thick black lines are 0
isogonal conjugates. Moreover, as we can see above, one direct consequence of the definition of an isogonal conjugate is that the gray angles are congruent, and the black angles are congruent. P Q Figure 0: Isogonal conjugate points. Since isogonal conjugates inherently involve angles, one question which naturally arises is how isogonal conjugates relate to triangles. s Theorem below states, they have at least one fascinating property. Theorem. Let P be a point in the plane, and let be given. Then the lines isogonal to P, P, and P, meet at a point Q. Points P and Q are called isogonal conjugate points, or isogonal conjugates. [Honsberger, pg. 53] proof of this theorem can be found in Ross Honsberger s work, Episodes in Nineteenth and Twentieth entury Euclidean Geometry. We will omit it here. However, Theorem is an important part of this section since it states that isogonal conjugate points, like isogonal conjugate lines, occur in pairs. Surprisingly, O and H, the circumcenter and orthocenter of a triangle, are one such pair. lthough this fact is well-known, our proof of it is included below. efore we begin, however, we must state a property of isogonal conjugates which we will use in the proof. F E D P Q Figure 1: Proposition 4. 1
Proposition 4. Let P and Q be isogonal conjugate lines through vertex, and let D be a point on P. Then the line EF connecting the feet of the perpendiculars from D to and is perpendicular to Q. [Honsberger, pgs. 64 65] proof of Proposition 4 can be found in Honsberger. We will omit it here and move directly to our proof that the circumcenter and the orthocenter are isogonal conjugates. Proposition 5. The circumcenter and the orthocenter of a triangle are isogonal conjugates. D P M b P a H a Figure : Isogonal conjugates: the circumcenter and the orthocenter. Proof. Let be given. onstruct H a, the altitude from vertex to. onstruct the isogonal conjugate of H a. It will meet at a point P a. Now, construct the perpendicular bisector of. It will meet P a at a point P. Drop a perpendicular from P to, and call the intersection of the two D. onstruct DM b, where M b is the midpoint of. y Proposition 4, DM b must be perpendicular to H a. However, by construction, H a is perpendicular to. D Thus, DM b. y similarity, DM b. So, = M b = 1. Therefore, D is the midpoint of, and P must be the circumcenter of. Thus, P lies on the isogonal conjugate to the altitude from vertex. y a similar argument, P will also lie on the isogonal conjugate to the altitude from vertices and. Therefore, the circumcenter and the orthocenter of are isogonal conjugates. P b P c P P a Figure 3: pedal triangle.
Returning to more general isogonal conjugates, another known and interesting property which we will reference later relates the circumcircles of the pedal triangles of two isogonal conjugate points. efore we state and prove this property, however, we need to define the concepts of a pedal triangle and a cyclic quadrilateral, and prove a useful fact about cyclic quadrilaterals. Definition 11. Let P be a point inside a given triangle. onstruct the perpendiculars from P to,, and. Label the points of intersection P c, P a, and P b, respectively. Then, P a P b P c is the pedal triangle of P with respect to. The next definition and lemma deal with certain types of quadrilaterals. Definition 1. quadrilateral D is called cyclic if all of its vertices lie on a circle. Lemma 1. Let D be a quadrilateral with right angles at vertices and. Then, quadrilateral D is cyclic. D M D Figure 4: generic cyclic quadrilateral and a cyclic quadrilateral with right angles. Proof. onnect to D. Then, the midpoint M of D will be the circumcenter of D and of D. Thus, M = MD = M = M. Thus, D lies on the circle centered at M, and D is cyclic. Now, let us state and prove the property of isogonal conjugates which relates the circumcircles of the pedal triangles of two isogonal conjugate points. Note that although this fact is known, the following proof is our own. Proposition 6. Let triangle be given with isogonal conjugate points P and Q. Then, the midpoint O of the segment P Q is the circumcenter of the pedal triangles of both P and Q. Moreover, both pedal triangles share the same circumcircle. [Honsberger, pgs. 67 69] Proof. The following proof utilizes Figure 5. Let P and Q be isogonal conjugate points, and let O be the midpoint of segment P Q. Let M be the midpoint of segment P, and let triangle P a P b P c be the pedal triangle determined by the point P. Let I be the intersection of segment P b P c with OM, and let J be the intersection of segment P b P c with Q. We first show that OM Q. y construction, MP = 1 P and OP = 1 P Q. Since MP O is common, it follows that MP O P Q. Thus, P MO P Q, which indicates that OM Q. 3
P c P a M I J P O Q L P b Figure 5: Proof that the pedal triangles of points P and Q share the same circumcenter. Now consider triangles P c P and P b L, where L is the intersection of P b P with Q. y construction, m P P c = π = m LP b. Moreover, since P and Q are isogonal conjugates, P c P = P b L. y angle subtraction, P c P = P b L. Thus, by Similarity, P c P P b L. Now, since m LP b = π = m P P c, quadrilateral P b P P c is cyclic by Lemma 1. Thus, P c P = P c P b P. Since m P c P + m P c P = P c P b P + m P b L = π, it follows that m P b JL = π. This implies that m P cio = π since OM Q. Now, since quadrilateral P b P P c is cyclic, the points, P b, P, and P c lie on a circle centered at M, the midpoint of P. Thus, MP c = MPb. Moreover, since segment MI is common, MP c I and MP b I both have right angles, congruent hypotenuses, and one congruent leg. They are consequently congruent. This then indicates that IP c = IPb, making I the midpoint of P b P c. Therefore, OM is the perpendicular bisector of P b P c. similar argument shows that O also lies on the perpendicular bisectors to segments P a P b and P a P c, making O the circumcenter of P a P b P c. Repeat the argument above for Q. Then, the midpoint O of segment P Q is the circumcenter of the pedal triangles constructed from P and Q. To show the two pedal triangles share not only the same circumcenter, but also the same circumcircle, consider the following argument, which utilizes Figure 6 below. Let points P and Q be isogonal conjugates, and let O be the midpoint of segment P Q. Drop perpendiculars from P, O, and Q to, and let P b, O b, and Q b be their intersections with, respectively. Since m P P b Q b = m OO b Q b = m QQ b O b = π, it follows that P P b OO b QQ b. onstruct P Q b and let N be the intersection of P Q b and OO b. Then, P NO = P Q b Q and P ON P O = P QQ b. Since NP O is common, P NO P Q b Q. Thus, P Q = P N P Q b = 1, and N is the midpoint of segment P Q b. y a similar argument, Q b O b N Q b P b P. Thus, = P bo b P b Q b = 1. This implies that O b is the midpoint of segment P b Q b. Thus, since P b O b = Pb Q b, P N P Q b P b O b O = Q b O b O, and OO b is common, by SS congruence, OO b P b = OOb Q b. This indicates that OP b = OQb. However, OP b is the radius of the pedal triangle of P, and OQ b is the radius of the pedal triangle of Q. Therefore, the pedal triangles of P and Q share the same circumcircle. Now that we have a basic understanding of isogonal conjugates and some of their properties, we are ready to define the symmedian point. 4
P O N Q P b O b Q b Figure 6: Proof that the pedal triangles of points P and Q share the same circumcircle. Definition 13. The symmedian point K is the isogonal conjugate of the centroid. Interesting properties of the symmedian include such statements as Proposition 7 below. Proposition 7. The symmedian point is the centroid of its own pedal triangle. [Honsberger, pgs. 7 73] This proposition is well known. proof of it can be found in Honsberger. Moreover, utilizing it and Proposition 6, we can immediately prove the following corollary. orollary. The centroid of a always lies on the Euler line of the pedal triangle of its symmedian point K. Proof. Let be given. Since the centroid and the circumcenter of a triangle determine its Euler line, it follows from Propositions 6 and 7 that GK, where G is the centroid of and K is its symmedian point, is the Euler line of the pedal triangle of K. orollary leads to an interesting observation. The locus of symmedian points in a gr -family of triangles form either an entire circle or an arc of a circle centered on the Euler line. Let us state this more formally. Theorem 3. Let Ω be the family of triangles produced by fixing a circumcenter O, a centroid G, a circumradius r, and varying GO from 0 to π. Let g denote the distance between O and G. Let E [0, π] such that for all φ E, there exists a triangle φ Ω with m GO = φ. Let K φ be the symmedian point of φ. Let K n be the point on the ray OG such that OKn = gr. Then, for all φ E, K r g φ lies on g the circle with radius r centered at K r g n. We call this circle the arleton ircle and its center the Knights Point. K n If g < r 3, every point on the arleton ircle is the symmedian point of a triangle in Ω. If g = r 3, then every point except (r, 0) is the symmedian point of a triangle in Ω. If g > r 3, then every point P on the arleton ircle such that P is strictly contained in the interior disc enclosed by the circumcircle of Ω is the symmedian point of a triangle in Ω. 5
While in the specific case in which g = r 3 a geometric proof of Theorem 3 is apparent, in the general case, such a proof is not obvious. Thus, we will approach the general case from an analytic perspective. However, in order to more clearly understand precisely what Theorem 3 states, let us begin our exploration with the geometric proof of the case in which g = r 3. Lemma. Let Ω be a gr family of triangles in which g = r 3. Let K n lie on segment OH such that OK n = 3K n H. Let the circle centered at K n with radius K n H be called the arleton ircle. Then, for every φ Ω, K φ, the symmedian point of φ, lies on the arleton ircle. Moreover, every point except H on the arleton ircle is the symmedian point of a triangle in the given Ω. Note that Lemma s geometric description of the location of K n and the radius of the arleton ircle correspond to the analytic description in Theorem 3. This is because by Proposition 7 and orollary 1, when g = r 3, H will coincide with one vertex of φ, making OH a radius of the circumcircle. Thus, K n will lie the following distance along the Euler line of : g(3g) (3g) g g (3g) (3g) g gr r g = = 18g3 = 9g 8g 4 = 3r 4 = 3 4 OH. Moreover, the radius of the arleton ircle will be g r = r g = 6g3 = 3g 8g 4 = r 4 = 1 4 OH. Now, to prove Lemma, we will use the following proposition. Proposition 8. Let be a right triangle with the right angle at vertex. Then, the symmedian point K is the midpoint of the symmedian from vertex. [Honsberger, pgs. 59 60] This fact is commonly known; a proof of it can be found in Honsberger. Let us now prove Lemma. K b J K ϕ O M K n Figure 7: Proof that the symmedian point lies on a circle when is a right triangle. Proof. y orollary 1 and Proposition 6, one vertex of φ must coincide with H. Without loss of generality, assume that is that vertex. onstruct the symmedian from, and let K b be the intersection of that symmedian with. We will first show that K b lies on the circle centered at M, the midpoint of segment OH. onstruct the angle bisector of and label it J, where J is its intersection with. Now, since the symmedian is the isogonal conjugate of the median, OJ = JK b and K b = O. Since O is the circumcenter of, it follows that O is isosceles with O = O. 6