Geometry. Class Examples (July 3) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

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1 Geometry lass Examples (July 3) Paul Yiu Department of Mathematics Florida tlantic University c b a Summer 2014

2 Example 11(a): Fermat point. Given triangle, construct externally similar isosceles triangles X, Y, and Z with base angles θ. 27 Z Y θ θ (a) pplying the law of cosines to triangle X, X 2 = 2 + X 2 2 X cos X ( a ) 2 = c 2 ca + cos( + θ) 2cosθ cos θ = c 2 + a2 4cos 2 θ = c 2 + a2 4cos 2 θ X ca(cos cos θ sin sin θ) cos θ ca cos + ca sin tan θ = c 2 + a2 4cos 2 θ 1 2 (c2 + a 2 b 2 )+2Δtanθ = 1 2 (a2 + b 2 + c 2 )+ 1 4cos2 θ a 2 +2Δtanθ. 4cos 2 θ (b) This expression is symmetric in a, b, c if and only if cos θ = 1 2, i.e., θ = ±60. With θ =60,wehave X 2 = 1 2 (a2 + b 2 + c Δ). This means that if equilateral triangles X, Y, and Z are constructed externally of triangle, the segments X, Y, Z have equal lengths.

3 28 Example 11(b): The Fermat point. Given triangle, construct equilateral triangles X, Y, and Z externally on the sides. Let the circumcircle of X intersect the line X at F. Y Z F X (a) Prove that FX = XF =60. (b) Prove that, F,, Y are concyclic. Similarly,, F,, Z are also concyclic. (c) Prove that, F, Y are collinear. Similarly,, F, Z are also collinear. The point F is called the Fermat point of triangle. It is the point of concurrency of the lines X, Y, Z. It is also the common point of the circumcircles of the equilateral triangles X, Y, Z.

4 29 pollonius Theorem Theorem. Given triangle, let D be the midpoint of. The length of the median D is given by =2(D 2 + D 2 ). D Proof. pplying the law of cosines to triangles D and D, and noting that cos D = cos D,wehave 2 = D 2 + D 2 2D D cos D; 2 = D 2 + D 2 2D D cos D, 2 = D 2 + D 2 +2D D cos D. The result follows by adding the first and the third lines. If m a denotes the length of the median on the side, m 2 a = 1 4 (2b2 +2c 2 a 2 ).

5 30 Example 12. is a triangle with a =8, b =9, c =11. Two of its medians have rational lengths. What are these? 8 D 9 E F mb = 17 2 ; m c = 13 2.

6 Example 13. The lengths of the sides of a triangle are 136, 170, and 174. alculate the lengths of its medians. 31

7 ngle bisector theorem Theorem (ngle bisector theorem). The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If X and X respectively the internal and external bisectors of angle, then X : X = c : b and X : X = c : b. Z c Z b X X Proof. onstruct lines through parallel to the bisectors X and X to intersect the line at Z and Z. (1) Note that Z = X = X = Z. This means Z =. learly, X : X = : Z = : = c : b. (2) Similarly, Z =, and X : X = : Z = : = c : b.

8 2 Example D1. square is inscribed in a right triangle with sides a and b. Show that each side of the square has length l = ab a + b. a l l b

9 3 Solution to Example D1. square is inscribed in a right triangle with sides a and b. Show that each side of the square has length l = ab a + b. P a l l b Solution. P + P = = 1 2 at bt = 1 2 ab = (a + b)t = ab = t = ab a+b.

10 4 Example D2. In triangle, α = 120. X is the bisector of angle. Show that 1 t = 1 b + 1 c. c t b X

11 5 Solution to Example D2. In triangle, α = 120. X is the bisector of angle. Show that 1 t = 1 b + 1 c. c t b X

12 6 Example D3. In the diagram below, X, Y, and DZ are equilateral triangles. Suppose XY Z = 120. Show that 1 b = 1 a + 1 c. Hint: Extend ZY to intersect at T. Show that T = a. Z X Y a b c D

13 7 Solution to Example D3. In the diagram below, X, Y, and DZ are equilateral triangles. Suppose XY Z = 120. Show that 1 b = 1 a + 1 c. Z X Y a T b c D Solution. Since XY T = XT =60, the points X, Y,, T are concyclic. Therefore, XTY = XY =60 and triangle XY T is equilateral. y Ptolemy s theorem, X = T + Y, and T = T + = T + Y = X = a. Now, triangle TZ has a 120 angle at with bisector Y = b. Therefore, 1 b = 1 a + 1 c.

14 10 Theorem. (a) The lengths of the internal and external bisectors of angle are respectively t a = 2bc b + c cos α 2 and t a = 2bc b c sin α 2. c t a b t a X X Proof. Let X and X be the bisectors of angle. (1) onsider the area of triangle as the sum of those of triangles X and X. Wehave 1 2 t a(b + c)sin α 2 = 1 bc sin α. 2 From this, t a = bc b + c sin α sin α 2 = 2bc b + c cos α 2. (2) onsider the area of triangle as the difference between those of X and X. Remarks. (1) 2bc b+c is the harmonic mean of b and c. It can be constructed as follows. If the perpendicular to X at X intersects and at Y and Z, then Y = Z = 2bc b+c. c t a b Z X Y

15 12 Example D5. The lengths of the sides of a triangle are 84, 125, 169. alculate the lengths of its internal bisectors. 1 1 nswers: 975 7, ,

16 Example D6. is a right triangle in which the bisector of the right angle, and the median to the hypotenuse have lengths 24 2 and 35 respectively. alculate the sidelengths of the triangle. 13

17 14 Example D7. Find an isosceles triangle for which the bisector of a base angle is the geometric mean of the two segments it divides on the opposite side. 2 Z Y 2 nswer: (a, b, c) =(1, 1+ 2, 1+ 2).

18 15 Example D8. In, α =60, and <. The bisector of intersects at X. If X is a mean proportional between X and X, find angle. X

19 16 Solution to Example D8. In, α =60, and <. The bisector of intersects at X. If X is a mean proportional between X and X, find angle. X Solution. y the angle bisector theorem, X = ac ab b+c, X = b+c. The length of the bisector X is 2bc b+c cos 2 = 3bc b+c since =60. If X is a mean proportional of X and X, then 3b 2 c 2 (b + c) = a2 bc 2 (b + c) 2. From this, a 2 =3bc. y the law of cosines, a 2 = b 2 + c 2 bc. Therefore, b 2 + c 2 bc =3bc, b 2 + c 2 4bc =0. Solving this equation, we obtain c =(2± 3)b. Since b<c,wehavec =(2+ 3)b. tan 2 tan + 2 c b c + b = = 1 3. This means = 1 3. ut + = 120 = tan + 2 = 3. It follows that tan 2 =1, and =90. Hence, = 105 and =15.

20 Example D. Suppose is a triangle with, and let D, E, X, Y be points on the line defined as follows: D is the midpoint of, E is the foot of the perpendicular from to, X, Y bisect angle. Prove that = DE XY. 17 Y E X D

21 18 The circle of pollonius Theorem. and are two fixed points. For a given positive number k 1, 1 the locus of points P satisfying P : P = k :1is the circle with diameter XY, where X and Y are points on the line such that X : X = k :1and Y : Y= k : 1. P Y O X Proof. Since k 1, points X and Y can be found on the line satisfying the above conditions. onsider a point P not on the line with P : P = k :1. Note that PX and PY are respectively the internal and external bisectors of angle P. This means that angle XPY is a right angle, and P lies on the circle with XY as diameter. onversely, let P be a point on this circle. We show that P : P = k : 1. Let be a point on the line such that PX bisects angle P. Since P and P are perpendicular to each other, the line P is the external bisector of angle P, and Y = X = X = Y X X On the other hand, Y Y = X X = X X Y X YX. = Y X YX. omparison of the two expressions shows that coincides with, and PX is the bisector of angle P. It follows that P P = X X = k. 1 If k =1, the locus is clearly the perpendicular bisector of the segment.

22 19 Example D9. If = d, and k 1, the radius of the pollonius circle is k k 2 1 d.

23 20 Example D10. Given two disjoint circles () and (), find the locus of the point P such that the angle between the pair of tangents from P to () and that between the pair of tangents from P to () are equal. P

24 21 Steiner-Lehmus theorem Lemma. Let triangles and XY Z be equiangular. If > XY, then > XZ and > Y Z. Z Given: Equiangular triangles and XYZ with > XY. To prove: > XZ and > Y Z. onstruction: Let be a point on such that = XY. onstruct the parallel through to, intersecting the line at. Proof : must be between and, for otherwise, and are on opposite sides of, and the lines, intersect. This contradicts their parallelism. Now, triangles and XY Z are congruent. [S] Therefore, > = XZ. X Y

25 22 Theorem. triangle is isosceles if and only if it has two equal internal angle bisectors. Given: Triangle with <and bisectors E, F. To prove: E > F. onstruction: Point G on such that GF = E. Join G and let it intersect F at H. G F H E Proof : The triangles GH and GF are equiangular with G > G. Therefore, H > F. (Lemma above) Since E > H, E > F.

Chapter 3. The angle bisectors. 3.1 The angle bisector theorem

Chapter 3. The angle bisectors. 3.1 The angle bisector theorem hapter 3 The angle bisectors 3.1 The angle bisector theorem Theorem 3.1 (ngle bisector theorem). The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If