Three Lemmas in Geometry

Winter amp 2010 Three Lemmas in Geometry Yufei Zhao Three Lemmas in Geometry Yufei Zhao Massachusetts Institute of Technology yufei.zhao@gmail.com 1 iameter of incircle T Lemma 1. Let the incircle of triangle touch side at, and let T be a diameter of the circle. If line T meets at, then =. F T E Z Y Proof. ssume wlog that. onsider the dilation with center that carries the incircle to an excircle. The line segment T is the diameter of the incircle that is perpendicular to, and therefore its image under the dilation must be the diameter of the excircle that is perpendicular to. It follows that T must get mapped to the point of tangency between the excircle and. In addition, the image of T must lie on the line T, and hence T gets mapped to. Thus, the excircle is tangent to at. It remains to prove that =. Let the incircle of touch sides and at F and E, respectively. Let the excircle of opposite to touch rays and at Z and Y, respectively, 1

Winter amp 2010 Three Lemmas in Geometry Yufei Zhao then using equal tangents, we have 2 = F + + = F + Z + = F Z + = EY + = E + Y + = + + = 2. Thus =. Problems 1. (IM 1992) In the plane let be a circle, l a line tangent to the circle, and M a point on l. Find the locus of all points P with the following property: there exists two points Q, R on l such that M is the midpoint of QR and is the inscribed circle of triangle P QR. 2. (USM 1999) Let be an isosceles trapezoid with. The inscribed circle ω of triangle meets at E. Let F be a point on the (internal) angle bisector of such that EF. Let the circumscribed circle of triangle F meet line at and G. Prove that the triangle F G is isosceles. 3. (IM Shortlist 2005) In a triangle satisfying + = 3 the incircle has centre I and touches the sides and at and E, respectively. Let K and L be the symmetric points of and E with respect to I. Prove that the quadrilateral KL is cyclic. 4. (Nagel line) Let be a triangle. Let the excircle of opposite to touch side at. Similarly define E on and F on. Then, E, F concur (why?) at a point N known as the Nagel point. Let G be the centroid of and I the incenter of. Show that I, G, N lie in that order on a line (known as the Nagel line, and GN = 2IG. 5. (USM 2001) Let be a triangle and let ω be its incircle. enote by 1 and E 1 the points where ω is tangent to sides and, respectively. enote by 2 and E 2 the points on sides and, respectively, such that 2 = 1 and E 2 = E 1, and denote by P the point of intersection of segments 2 and E 2. ircle ω intersects segment 2 at two points, the closer of which to the vertex is denoted by Q. Prove that Q = 2 P. 6. (Tournament of Towns 2003 Fall) Triangle has orthocenter H, incenter I and circumcenter. Let K be the point where the incircle touches. If I is parallel to, then prove that is parallel to HK. 7. (IM 2008) Let be a convex quadrilateral with. enote the incircles of triangles and by ω 1 and ω 2 respectively. Suppose that there exists a circle ω tangent to the ray beyond and to the ray beyond, which is also tangent to the lines and. Prove that the common external tangents of ω 1 and ω 2 intersect on ω. (Hint: show that + = +. What does this say about the lengths along?) 2

Winter amp 2010 Three Lemmas in Geometry Yufei Zhao 2 enter of spiral similarity spiral similarity 1 about a point (known as the center of the spiral similarity) is a composition of a rotation and a dilation, both centered at. (See diagram) Figure 1: n example of a spiral similarity. For instance, in the complex plane, if = 0, then spiral similarities are described by multiplication by a nonzero complex number. That is, spiral similarities have the form z αz, where α \{0}. Here α is the dilation factor, and arg α is the angle of rotation. It is easy to deduce from here that if the center of the spiral similarity is some other point, say z 0, then the transformation is given by z z 0 + α(z z 0 ) (why?). Fact. Let,,, be four distinct point in the plane such that is not a parallelogram. Then there exists a unique spiral similarity that sends to, and to. Proof. Let a, b, c, d be the corresponding complex numbers for the points,,,. We know that a spiral similarity has the form T(z) = z 0 + α(z z 0 ), where z 0 is the center of the spiral similarity, and α is data on the rotation and dilation. So we would like to find α and z 0 such that T(a) = c and T(b) = d. This amount to solving the system Solving it, we see that the unique solution is z 0 + α(a z 0 ) = c, z 0 + α(b z 0 ) = d. α = c d a b, z 0 = ad bc a b c + d. Since is not a parallelogram, a b c + d 0, so that this is the unique solution to the system. Hence there exists a unique spiral similarity that carries to and to. Exercise. How can you quickly determine the value of α in the above proof without even needing to set up the system of equations? Exercise. Give a geometric argument why the spiral similarity, if it exists, must be unique. (Hint: suppose that T 1 and T 2 are two such spiral similarities, then what can you say about T 1 T 1 2?) So we know that a spiral similarity exists, but where is its center? The following lemma tells us how to locate it. 1 If you want to impress your friends with your mathematical vocabulary, a spiral similarity is sometimes called a similitude, and a dilation is sometimes called a homothety. (ctually, they are not quite exactly the same thing, but shhh!) 3

Winter amp 2010 Three Lemmas in Geometry Yufei Zhao Lemma 2. Let,,, be four distinct point in the plane, such that is not parallel to. Let lines and meet at. Let the circumcircles of and meet again at. Then is the center of the unique spiral similarity that carries to and to. Proof. We use directed angles mod π (i.e., directed angles between lines, as opposed to rays) in order to produce a single proof that works in all configurations. Let (l 1, l 2 ) denote the angle of rotation that takes line l 1 to l 2. useful fact is that points P, Q, R, S are concyclic if and only if (P Q, QR) = (P S, SR). We have (, ) = (, ) = (, ) = (, ), and (, ) = (, ) = (, ) = (, ). It follows that triangles and are similar and have the same orientation. Therefore, the spiral similarity centered at that carries to must also carry to. Finally, it is worth mentioning that spiral similarities come in pairs. If we can send to, then we can just as easily send to using the same center. Fact. If is the center of the spiral similarity that sends to and to, then is also the center of the spiral similarity that sends to and to. Proof. Since spiral similarity preserves angles at, we have =. lso, the dilation ratio of the first spiral similarity is / = /. So the rotation about with angle = followed by a dilation with ratio / = / sends to, and to, as desired. Problems 1. (IM Shortlist 2006) Let E be a convex pentagon such that = = E and = = E. iagonals and E meet at P. Prove that line P bisects side. 2. (USM 2006) Let be a quadrilateral, and let E and F be points on sides and, respectively, such that E/E = F/F. Ray F E meets rays and at S and T, respectively. Prove that the circumcircles of triangles SE, SF, T F, and T E pass through a common point. 3. (hina 1992) onvex quadrilateral is inscribed in circle ω with center. iagonals and meet at P. The circumcircles of triangles P and P meet at P and Q. ssume that points, P, and Q are distinct. Prove that QP = 90. 4

Winter amp 2010 Three Lemmas in Geometry Yufei Zhao 4. Let be a quadrilateral. Let diagonals and meet at P. Let 1 and 2 be the circumcenters of P and P. Let M, N and be the midpoints of, and 1 2. Show that is the circumcenter of MP N. 5. (Miquel point of a quadrilateral) Let l 1, l 2, l 3, l 4 be four lines in the plane, no two parallel. Let ijk denote the circumcircle of the triangle formed by the lines l i, l j, l k (these circles are called Miquel circles). Then 123, 124, 134, 234 pass through a common point (called the Miquel point). (It s not too hard to prove this result using angle chasing, but can you see why it s almost an immediate consequence of the lemma?) 6. (IM 2005) Let be a given convex quadrilateral with sides and equal in length and not parallel. Let E and F be interior points of the sides and respectively such that E = F. The lines and meet at P, the lines and EF meet at Q, the lines EF and meet at R. onsider all the triangles P QR as E and F vary. Show that the circumcircles of these triangles have a common point other than P. 7. (IM Shortlist 2006) Points 1, 1 and 1 are chosen on sides,, and of a triangle, respectively. The circumcircles of triangles 1 1, 1 1, and 1 1 intersect the circumcircle of triangle again at points 2, 2, and 2, respectively ( 2, 2, and 2 ). Points 3, 3, and 3 are symmetric to 1, 1, 1 with respect to the midpoints of sides,, and, respectively. Prove that triangles 2 2 2 and 3 3 3 are similar. 3 Symmedian Let be a triangle. Let M be the midpoint of, so that M is a median of. Let N be a point on side so that M = N. Then N is a symmedian of. In other words, the symmedian is the reflection of the median across the angle bisector. The following lemma gives an important property of the symmedian. Lemma 3. Let be a triangle and Γ its circumcircle. Let the tangent to Γ at and meet at. Then coincides with a symmedian of. We give three proofs. (The three diagram each correspond to a separate proof.) The first proof is a sine law chase. 5

Winter amp 2010 Three Lemmas in Geometry Yufei Zhao M' P M F E Q First proof. Let the reflection of across the angle bisector of meet at M. Then M M = M sin M sin M sin M sin sin M sin = sin sin M sin sin = sin sin = = 1. [Sine law on M and M ] [Using the tangent-chord angles] [From construction of M ] [Sine law on and ] Therefore, M is the median, and thus is the symmedian. Remark. Some people like to start this proof by setting M to be the midpoint of, and then using sin sin M sine law to show that sin = sin M. I do not recommend this variation, since it s not immediately clear that M = follows, especially when is obtuse. Next we give a synthetic proof that highlights some additional features in the configuration. Second proof. Let be the circumcenter of and let ω be the circle centered at with radius. Let lines and meet ω at P and Q, respectively. Since = QP, triangles and QP are similar. The idea is use the fact that, up to dilation, triangles and QP are reflections of each other across the angle bisector of. Since P Q = Q + = 1 2 ( + ) = 90, we see that P Q is a diameter of ω and hence passes through. Let M be the midpoint of. Since is the midpoint of QP, the similarity implies that M = Q, from which the result follows. The third proof uses facts from projective geometry. Feel free to skip it if you are not comfortable with projective geometry. Third proof. Let the tangent of Γ at meet line at E. Then E is the pole of (since the polar of is E and the pole of is ). Let meet at F. Then point,, E, F are harmonic. This means that line,, E, F are harmonic. onsider the reflections of the four line across the angle bisector of. Their images must be harmonic too. It s easy to check that E maps onto a line 6

Winter amp 2010 Three Lemmas in Geometry Yufei Zhao parallel to. Since must meet these four lines at harmonic points, it follows that the reflection of F must pass through the midpoint of. Therefore, F is a symmedian. Problems 1. (Poland 2000) Let be a triangle with =, and P a point inside the triangle such that P = P. If M is the midpoint of, then show that P M + P = 180. 2. (IM Shortlist 2003) Three distinct points,, are fixed on a line in this order. Let Γ be a circle passing through and whose center does not lie on the line. enote by P the intersection of the tangents to Γ at and. Suppose Γ meets the segment P at Q. Prove that the intersection of the bisector of Q and the line does not depend on the choice of Γ. 3. (Vietnam TST 2001) In the plane, two circles intersect at and, and a common tangent intersects the circles at P and Q. Let the tangents at P and Q to the circumcircle of triangle P Q intersect at S, and let H be the reflection of across the line P Q. Prove that the points, S, and H are collinear. 4. (US TST 2007) Triangle is inscribed in circle ω. The tangent lines to ω at and meet at T. Point S lies on ray such that S T. Points 1 and 1 lies on ray ST (with 1 in between 1 and S) such that 1 T = T = 1 T. Prove that triangles and 1 1 are similar to each other. 5. Let be a triangle. Let be the center of spiral similarity that takes to and to. Show that coincides with a symmedian of. 6. (US TST 2008) Let be a triangle with G as its centroid. Let P be a variable point on segment. Points Q and R lie on sides and respectively, such that P Q and P R. Prove that, as P varies along segment, the circumcircle of triangle QR passes through a fixed point such that G =. 7. (US 2008) Let be an acute, scalene triangle, and let M, N, and P be the midpoints of,, and, respectively. Let the perpendicular bisectors of and intersect ray M in points and E respectively, and let lines and E intersect in point F, inside of triangle. Prove that points, N, F, and P all lie on one circle. 8. Let be one of the intersection points of circles ω 1, ω 2 with centers 1, 2. The line l is tangent to ω 1, ω 2 at, respectively. Let 3 be the circumcenter of triangle. Let be a point such that is the midpoint of 3. Let M be the midpoint of 1 2. Prove that 1 M = 2. (Hint: use Problem 5.) 7