A quick introduction to (Ceva s and) Menelaus s Theorem

Transcription

1 quick introduction to (eva s and) Menelaus s Theorem 1 Introduction Michael Tang May 17, 2015 Menelaus s Theorem, often partnered with eva s Theorem, is a geometric result that determines when three points related to a given triangle can be collinear. It is also related to the physics-mathematical technique of mass points; both are used to determine ratios of side lengths in triangles. Here we will examine the interesting statement and proof of Menelaus s Theorem, and give several example problems to illustrate its uses, both in computational and proof-based geometry. asic knowledge of eva s Theorem is assumed, but a quick refresher is also given. 2 Menelaus s Theorem efore we get to the full statement of Menelaus s Theorem, we will briefly introduce a convention that helps to generalize geometric proofs and theorems, including Menelaus. This convention is directed segments: like directed angles, in which angles have signs depending on their orientations (see [5] for more details on directed angles), but with segments on a line. When we use directed segments on a certain line l, lengths of segments in l that go in opposite directions are given opposite signs, and lengths of segments with the same direction have the same sign. This generalizes many theorems, most notably, a very basic and fundamental one: using directed segments, for any three collinear points,,, + =, regardless of whether or not is between and. (If is not between and, then and have opposite signs, so the left-hand side becomes either or, both of which equal in the appropriate circumstances.) Figure 1: Our length equalities still hold in this case! s seen, and as noted in [4], the concept of directed segments shares many similarities with vectors, just only in a single, given dimension. More relevant to Menelaus s Theorem, when we multiply or divide lengths on l, the result is positive if the segments have the same direction, and negative if they have opposite directions. 1

2 Now, the directed-segments statement of Menelaus s Theorem is as follows: Theorem 1 (Menelaus, Directed Segments). Direct all segments. In, let,, and be points on,, and, possibly extended. Then points,, are collinear if and only if = 1. To derive a non-directed-segments version of this theorem (which is more useful if there are no configuration issues, or if the problem is computational, not proof-based), notice, by the properties of directed segments, that / has a negative sign if and only if lies on the extension of segment. Therefore, if the equality in Theorem 1 holds, then exactly one or three of,, must lie on the extensions of their respective sides, so that the left-hand side can be negative. (This makes sense: if exactly two of,, are on extensions or they are all on their sides, it is impossible for,, to be collinear.) This gives us the following non-directed version of Menelaus, whose form is still very similar. (Essentially, we add a somewhat inconvenient configuration restriction, but then are allowed to drop the signs on our segments.) Figure 2: The two different configurations of Menelaus. Theorem 2 (Menelaus, Undirected Segments). In, let,, and be points on,, and, possibly extended. Then points,, are collinear if and only if exactly one or three of them lie on the extensions of their respective sides, and = 1. Menelaus s Theorem can be seen as a sort of counterpart to eva s Theorem, which deals with the concurrence of three cevians on a triangle: Theorem 3 (eva). Let,, be points on the sides,, of. Then segments,, are concurrent if and only if = 1. 2

3 oth deal with points on triangles: eva tells us when three lines (cevians) are concurrent, and Menelaus tells us when three points are collinear. Indeed, the form of the ratios in both results are exactly the same, which can help in remembering Menelaus if you already know eva, or vice versa. Following [1], Menelaus and eva can be seen as a sort of duality: if points,, satisfying the ratio condition are all on the sides of, then eva holds, and,, concur. ut if,, satisfying the same ratio condition are not all on the sides - specifically, one or three of them are on extensions of the sides - then Menelaus holds, and,, are collinear. It is indeed a beautiful result that the same ratio equality gives rise to both concurrence and collinearity, depending on the configuration at hand. 3 Proof of Menelaus s Theorem Here we prove the directed version of Menelaus s Theorem (and per our explanation from earlier, this will also imply the non-directed version). The standard proof of Menelaus s Theorem (as given in [4]) is, somewhat strangely, to drop perpendiculars. That seems unexpected, but it creates helpful similar triangles, allowing us to get at our ratios. First, we prove that collinearity implies the ratio condition. Suppose that,, are collinear, all lying on line l. Then, drop altitudes from,, to l, and call their lengths h a, h b, h c respectively. To deal with different configurations, we can give signs to h a, h b, h c as well: they are positive if they lie on one side of l, and negative otherwise. h a h c h a h b h c h b Figure 3: Dropping perpendiculars: the two possible configurations. Then, we have similar right triangles, giving (note the sign!) = h b h c, = h c h a, = h a h b. (ou can check that the signs work out in both configurations, given above.) Then multiplying, we have = h b h c h a = 1, h c h a h b as we wanted. 3

4 For the reverse direction, ratio condition implying collinearity, a simple application of phantom points is enough. Suppose that the ratio condition holds, and let line intersect at. Then, by Menelaus s Theorem with the points,,, = 1. ut, because the ratio condition holds with,,, we also have = 1. omparing the ratio conditions gives / = /, which is enough to imply that = : write = and =, remembering that these both hold in directed segments, no matter the configuration. Then we have ( )/ = ( )/, and adding one to both sides gives / = /, so =. y similarly writing and and manipulating, we can get =. These two conditions mean that =, regardless of whether, are on the side or an extension. 4 n example application Most basic applications of eva and Menelaus are just that: applying the two results when you want to prove collinearity/concurrence, or find ratios given collinearity/concurrence. Regular geometric reasoning usually takes over from there. So, with that in mind, let s look at an example problem. Example 1. (David ltizio) Triangle has = 2007 and = The incircle ω of the triangle is tangent to and at E and F respectively, and P is the intersection point of EF and. Suppose is the midpoint of P. ompute the length. Solution. This configuration is exactly that of Menelaus: P, F, E are the three collinear points, we are given essentially the ratio P/P, and we can use some basic properties of the incircle to find some other ratios. F E P Let a, b, c be the side lengths,, and s = (a + b + c)/2 be the semiperimeter, as usual. (We are given c = 2007, b = 2015, and we want to find a.) Then, it is a well known fact that F = E = s a, F = s b, and E = s c. (See the first part of [2] for a proof of this.) Therefore, by Menelaus s Theorem, P P F F E E = 1 = 2 s b s a s a s c = 1. 4

5 This simplifies to 2(s b) = s c. Then 2s 2b = s c, so s = 2b c. Hence (a + b + c)/2 = 2b c, and solving for a, we get a = 3b 3c. With c = 2007 and b = 2015, we find that a = 3( ) = 24. Of course, not all problems involving Menelaus will be as straightforward as this one, as you can see in the practice problems below! 5 Practice Problems Here are a few practice problems, both computational and proof-based. They are given in (very) rough order of increasing difficulty - rough being because computational and proofbased problems can be somewhat unorderable in terms of difficulty. No guarantees that all problems involve Menelaus directly, but they are all very interesting! 1. In, let E and F lie on sides and, respectively, so that lines EF and intersect at K, where is between and K. Let E and F intersect at P, and line P intersect at D. Prove that D D = K. (ops forums) K 2. Given noncollinear points,,, segment is trisected by points D and E, and F is the midpoint of segment. DF and F intersect E at G and H, respectively. If [DEG] = 18, compute [F GH]. (RML, 2012) H G F D E 3. The diagram below shows equilateral with side length 2. Point D lies on ray so that D = 4. Points E and F lie on and, respectively, so that E, F, and D are collinear, and the area of EF is half of the area of. ompute E F. (Purple omet, 2014) E F D 4. (From [4]) In scalene, show that the points at which the external bisectors of,, meet,,, respectively, are collinear. (The external bisector of is the line perpendicular to its internal bisector.) 5

6 5. (From [3]) Let have = 6 and = 3. Point E is such that E = 1 and E = 5. onstruct point F on segment such that E = F. EF and are extended to meet at D. If [EF ] [F D] = m n where m and n are positive integers, find m + n (note: [] denotes the area of ). 6. (From [6]) The diagonals and D of convex quadrilateral D meet at point M in such a way that M = M and DM = 2M. Suppose that and are points on M and respectively such that M = = 3. Show that the points D,, and are collinear. 7. (lanchet s Theorem) In acute triangle, F is the foot of the perpendicular from to, and P is a point on F. Let lines P and P meet and at D and E, respectively. Show that EF = DF. 6 On the Horizon This article was meant to be a quick introduction to Menelaus s Theorem (and eva s Theorem), set in a mostly computational context. ut Menelaus and eva fit into a larger model, that of projective geometry. In fact, techniques from projective geometry can be used to quickly prove eva and Menelaus at the same time! Further research into that field is definitely recommended, both for exploration or beauty s sake, and for occasional use on olympiads. Thanks very much to David ltizio for proofreading this article and giving some suggestions, as well as contributing an example problem! References [1] lexander ogmolny The Menelaus Theorem (1999), available at cut-the-knot.org/ 4travelers/evandMenelaus.shtml [2] lexander ogmolny, Incircles and Excircles in a Triangle, available at cut-the-knot.org/triangle/inexircles.shtml [3] ltheman Mock IME (2007), available at aops.com/wiki/index.php/ Mock_IME_1_ _Problems/Problem_7 [4] oxeter and Greitzer Geometry Revisited 1967: The Mathematical ssociation of merica, Inc., Washington, D.. [5] Evan hen How to Use Directed ngles (2015), available at mit.edu/~evanchen/ handouts/directed-ngles/directed-ngles.pdf [6] Hang Kim Hoo, Koh Khee Meng em On Menelaus Theorem (1996), available at 20Theorem%20(Hang%20Kim%20Hoo%20and%20Koh%20KM).pdf 6