Lemmas in Euclidean Geometry 1

Transcription

1 Lemmas in uclidean Geometry 1 Yufei Zhao yufeiz@mit.edu 1. onstruction of the symmedian. Let be a triangle and Γ its circumcircle. Let the tangent to Γ at and meet at. Then coincides with a symmedian of. (The symmedian is the reflection of the median across the angle bisector, all through the same vertex.) O ' P Q We give three proofs. The first proof is a straightforward computation using Sine Law. The second proof uses similar triangles. The third proof uses projective geometry. irst proof. Let the reflection of across the angle bisector of meet at. Then = sin sin sin sin = sin sin sin sin = sin sin sin sin = = 1 Therefore, is the median, and thus is the symmedian. Second proof. LetO bethecircumcenter of andlet ω bethecircle centered at with radius. Let lines and meet ω at P and Q, respectively. Since PQ = Q+ = 1 2 ( + O) = 90, we see that PQ is a diameter of ω and hence passes through. Since = QP and = PQ, we see that triangles and QP are similar. f is the midpoint of, noting that is the midpoint of QP, the similarity implies that = Q, from which the result follows. Third proof. Let the tangent of Γ at meet line at. Then is the pole of (since the polar of is and the pole of is ). Let meet at. Then point,,, are harmonic. This means that line,,, are harmonic. onsider the reflections of the four line across the angle bisector of. Their images must be harmonic too. t s easy to check that maps onto a line parallel to. Since must meet these four lines at harmonic points, it follows that the reflection of must pass through the midpoint of. Therefore, is a symmedian. 1 Updated July 20,

2 (i) (Poland 2000) Let be a triangle with =, and P a point inside the triangle such that P = P. f isthemidpointof, thenshowthat P+ P = 180. (ii) (O 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. (iii) (Vietnam TST 2001) n 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 PQ intersect at S, and let H be the reflection of across the line PQ. Prove that the points, S, and H are collinear. (iv) (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. (v) (US 2008) Let be an acute, scalene triangle, and let, N, and P be the midpoints of,, and, respectively. Let the perpendicular bisectors of and intersect ray in points and respectively, and let lines and intersect in point, inside of triangle. Prove that points, N,, and P all lie on one circle. 2. iameter of the incircle. Let the incircle of triangle touch side at, and let be a diameter of the circle. f line meets at, then =. Proof. onsider the dilation with center that carries the incircle to an excircle. The diameter of the incircle must be mapped to the diameter of the excircle that is perpendicular to. t follows that must get mapped to the point of tangency between the excircle and. Since the image of must lie on the line, it must be. That is, the excircle is tangent to at. Then, it follows easily that =. (i) (O Shortlist 2005) n a triangle satisfying + = 3 the incircle has centre and touches the sides and at and, respectively. Let K and L be the symmetric points of and with respect to. Prove that the quadrilateral KL is cyclic. 2

3 (ii) (O 1992) n the plane let be a circle, l a line tangent to the circle, and a point on l. ind the locus of all points P with the following property: there exists two points Q,R on l such that is the midpoint of QR and is the inscribed circle of triangle PQR. (iii) (USO 1999) Let be an isosceles trapezoid with. The inscribed circle ω of triangle meets at. Let be a point on the (internal) angle bisector of such that. Let the circumscribed circle of triangle meet line at and G. Prove that the triangle G is isosceles. (iv) (USO 2001) Let be a triangle and let ω be its incircle. enote by 1 and 1 the points whereω is tangent tosides and, respectively. enote by 2 and 2 thepoints on sides and, respectively, such that 2 = 1 and 2 = 1, and denote by P the point of intersection of segments 2 and 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. (v) (Tournament of Towns 2003 all) Triangle has orthocenter H, incenter and circumcenter O. Let K be the point where the incircle touches. f O is parallel to, then prove that O is parallel to HK. 3. ude, where s my spiral center? Let and be two segments, and let lines and meet at X. Let the circumcircles of X and X meet again at O. Then O is the center of the spiral similarity that carries to. X O Proof. Since OX and XO are cyclic, we have O = O and O = O. t follows that triangles O and O are similar. The result is immediate. Remember that spiral similarities always come in pairs: if there is a spiral similarity that carries to, then there is one that carries to. (i) (O Shortlist 2006) Let be a convex pentagon such that = = and = =. iagonals and meet at P. Prove that line P bisects side. (ii) (hina 1992) onvex quadrilateral is inscribed in circle ω with center O. iagonals and meet at P. The circumcircles of triangles P and P meet at P and Q. ssume that points O,P, and Q are distinct. Prove that OQP = 90. (iii) Let be a quadrilateral. Let diagonals and meet at P. Let O 1 and O 2 be the circumcenters of P and P. Let, N and O be the midpoints of, and O 1 O 2. Show that O is the circumcenter of PN. 3

4 (iv) (USO 2006) Let be a quadrilateral, and let and be points on sides and, respectively, such that / = /. Ray meets rays and at S and T, respectively. Prove that the circumcircles of triangles S, S, T, and T pass through a common point. (v) (O 2005) Let be a given convex quadrilateral with sides and equal in length and not parallel. Let and be interior points of the sides and respectively such that =. The lines and meet at P, the lines and meet at Q, the lines and meet at R. onsider all the triangles PQR as and vary. Show that the circumcircles of these triangles have a common point other than P. (vi) (O Shortlist 2002) ircles S 1 and S 2 intersect at points P and Q. istinct points 1 and 1 (not at P or Q) are selected on S 1. The lines 1 P and 1 P meet S 2 again at 2 and 2 respectively, and the lines 1 1 and 2 2 meet at. Prove that, as 1 and 1 vary, the circumcentres of triangles 1 2 all lie on one fixed circle. (vii) (US TST 2006) n acute triangle, segments,, and are its altitudes, and H is its orthocenter. ircle ω, centered at O, passes through and H and intersects sides and again at Q and P (other than ), respectively. The circumcircle of triangle OPQ is tangent to segment at R. Prove that R/R = /. (viii) (O Shortlist2006) Points 1, 1 and 1 arechosen onsides,, 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 and are similar. 4. rc midpoints are equidistant to vertices and in/excenters Let be a triangle, its incenter, and,, its excenters. On the circumcircle of, let be the midpoint of the arc not containing and let N be the midpoint of the arc containing. Then = = = and N = N = N = N. N Proof. Straightforward angle-chasing (do it yourself!). nother perspective is to consider the circumcircle of as the nine-point-circle of. 4

5 (i) (PO2007) Let beanacuteangledtrianglewith = 60 and >. Let be the incenter, and H the orthocenter of the triangle. Prove that 2 H = 3. (ii) (O 2006) Let be a triangle with incentre. point P in the interior of the triangle satisfies P+ P = P+ P. Show that P, and that equality holds if and only if P =. (iii) (Romanian TST1996) Let beacyclic quadrilateral and let bethe set of incenters and excenters of thetriangles,,, (16 points in total). Prove that there are two sets K and L of four parallel lines each, such that every line in K L contains exactly four points of. 5. is the midpoint of the touch-chord of the mixtilinear incircles Let be a triangle and its incenter. Let Γ be the circle tangent to sides,, as well as the circumcircle of. Let Γ touch and at X and Y, respectively. Then is the midpoint of XY. Q Y P Y X X T Proof. Let the point of tangency between the two circles be T. xtend TX and TY to meet the circumcircle of again at P and Q respectively. Note that P and Q are the midpoint of the arcs and. pply Pascal s theorem to PTQ and we see that X,,Y are collinear. Since lies on the angle bisector of XY and X = Y, must be the midpoint of XY. (i) (O 1978) n triangle, =. circle is tangent internally to the circumcircle of triangle and also to sides, at P,Q, respectively. Prove that the midpoint of segment PQ is the center of the incircle of triangle. (ii) Let be a triangle. ircle ω is tangent to and, and internally tangent to the circumcircle of triangle. The circumcircle and ω are tangent at P. Let be the incircle of triangle. Line P meets the circumcircle of at P and Q. Prove that Q = Q. 6. ore curvilinear incircles. ( generalization of the previous lemma) Let be a triangle, its incenter and a point on. onsider the circle that is tangent to the circumcircle of but is also tangent to, at, respectively. Then, and are collinear. 5

6 K /' Proof. There is a computational proof using asey s theorem 2 and transversal theorem 3. You can try to work that out yourself. Here, we show a clever but difficult synthetic proof (communicated to me via Oleg Golberg). enote Ω the circumcircle of and Γ the circle tangent tangent to the circumcircle of and lines,. Let Ω and Γ touch at K. Let be the midpoint of arc on Ω not containing K. Then K,, are collinear (think: dilation with center K carrying Γ to Ω). lso,,, are collinear, and =. Let line meet Γ again at. t suffices to show that is tangent to Γ. Note that K is subtended by K in Γ and K is subtended by K in Ω. Since K and K are homothetic with center K, we have K = K, implying that,k,, are concyclic. We have = = K. So K. Hence 2 = K. Since =, we have 2 = K, implying that K. Therefore, K = K = K (since,k,, are concyclic). Therefore, is tangent to Ω and the proof is complete. (i) (ulgaria 2005) onsider two circles k 1,k 2 touching externally at point T. line touches k 2 at point X and intersects k 1 at points and. Let S be the second intersection point of k 1 with the line XT. On the arc TS not containing and is chosen a point. Let Y be the tangent line to k 2 with Y k 2, such that the segment Y does not intersect the segment ST. f = XY S. Prove that: (a) the points,t,y, are concyclic. 2 asey s theorem, also known as Generalized Ptolemy Theorem, states that if there are four circles Γ 1,Γ 2,Γ 3,Γ 4 (could be degenerated into a point) all touching a circle Γ such that their tangency points follow that order around the circle, then t 12t 34 +t 23t 14 = t 13t 24, where t 12 is the length of the common tangent between Γ i and Γ j (if Γ i and Γ j on the same side of Γ, then take their common external tangent, else take their common internal tangent.) think the converse is also true if both equations hold, then there is some circle tangent to all four circles. 3 The transversal theorem is a criterion for collinearity. t states that if,, are three collinear points, and P is a point not on the line, and,, are arbitrary points on lines P,P,P respectively, then,, are collinear if and only if P P P + + P P P = 0, where the lengths are directed. n my opinion, it s much easier to remember the proof than to memorize this huge formula. The simplest derivation is based on relationships between the areas of [P],[P ], etc. 6

7 (b) is the excenter of triangle with respect to the side. (ii) (Sawayama-Thébault 4 ) Let be a triangle with incenter. Let a point on side. Let P be the center of the circle that touches segments,, and the circumcircle of, and let Q be the center of the circle that touches segments,, and the circumcircle of. Show that P,Q, are collinear. (iii) Let P be a quadrilateral inscribed in a circle Ω, and let Q be the quadrilateral formed by the centers of the fourcircles internally touching O and each of the two diagonals of P. Show that the incenters of the four triangles having for sides the sides and diagonals of P form a rectangle R inscribed in Q. (iv) (Romania 1997) Let be a triangle with circumcircle Ω, and a point on the side. Show that the circle tangent to Ω, and, and the circle tangent to Ω, and, are tangent to each other if and only if =. (v) (Romania TST 2006) Let be an acute triangle with. Let be the foot of the altitude from and ω the circumcircle of the triangle. Let ω 1 be the circle tangent to, and ω. Let ω 2 be the circle tangent to, and ω. Let l be the interior common tangent to both ω 1 and ω 2, different from. Prove that l passes through the midpoint of if and only if 2 = +. (vi) ( 10368) or each point O on diameter of a circle, perform the following construction. Let the perpendicular to at O meet the circle at point P. nscribe circles in the figures bounded by the circle and the lines and OP. Let R and S be the points at which the two incircles to the curvilinear triangles OP and OP are tangent to the diameter. Show that RPS is independent of the position of O. 7. oncurrent lines from the incircle. Let the incircle of touch sides,, at,, respectively. Let be the incenter of and be the midpoint of. Then the lines, and are concurrent. P N Q Proof. Let lines and meet at N. onstruct a line through N parallel to, and let it meet sides and at P and Q, respectively. We need to show that,n, are collinear, so it suffices to show that N is the midpoint of PQ. We present two ways to finish this off, one using Simson s line, and the other using spiral similarities. Simson line method: onsider the triangle P Q. The projections of the point onto the three sides of PQ are,n,, which are collinear, must lie on the circumcircle of PQ by Simson s theorem. ut since is an angle bisector, P = Q, thus PN = QN. 4 bit of history: this problem was posed by rench geometer Victor Thébault ( ) in the merican athematical onthly in 1938 (Problem 2887, 45 (1938) ) and it remained unsolved until However, in 2003, Jean-Louis yme discovered that this problem was independently proposed and solved by instructor Y. Sawayama of the entral ilitary School of Tokyo in 1905! or more discussion, see yme s paper at 7

8 Spiral similarity method: Note that P, N,, are concyclic, so = QP. Similarly, PQ =. So triangles PQ and are similar. Since =, we have P = Q, and thus PN = QN. (c.f. Lemma 3) (i) (hina 1999) n triangle,. Let be the midpoint of side, and let be a point on median. Let be the foot of perpendicular from to side, and let P be a point on segment. Let and N be the feet of perpendiculars from P to sides and, respectively. Prove that,, and N are collinear if and only if P = P. (ii) (O Shortlist 2005) The median of a triangle intersects its incircle ω at K and L. The lines through K and L parallel to intersect ω again at X and Y. The lines X and Y intersect at P and Q. Prove that P = Q. 8. ore circles around the incircle. Let be the incenter of triangle, and let its incircle touch sides,, at, and, respectively. Let line meet at T. Then T,,,, are concyclic. onsequent results include: T = 90, and T lies on the line connecting the midpoints of and. n easier way to remember the third part of the lemma is: for a triangle, draw a midline, an angle bisector, and a touch-chord, each generated from different vertex, then the three lines are concurrent. T Proof. Showing that,t,, are concyclic is simply angle chasing (e.g. show that = ). The second part follows from T = T = = 90. or the third part, note that if is the midpoint of, then is the midpoint of an hypotenuse of the right triangle T. So T =. Then T = T = T, so T is parallel to, and so T is a midline of the triangle. (i) Let be an acute triangle whose incircle touches sides and at and, respectively. Let the angle bisectors of and meet at X and Y, respectively, and let the midpoint of be Z. Show that XYZ is equilateral if and only if = 60. (ii) (O Shortlist 2004) or a given triangle, let X be a variable point on the line such that lies between and X and the incircles of thetriangles X and X intersect at two distinct points P and Q. Prove that the line PQ passes through a point independent of X. (iii) Let points and lie on the circle Γ, and let be a point inside the circle. Suppose that ω is a circle tangent to segments, and Γ. Let ω touch and Γ at P and Q. Show that the circumcircle of PQ passes through the incenter of. 8

9 9. Reflections of the orthocenter lie on the circumcircle. Let H be the orthocenter of triangle. Let the reflection of H across the be X and the reflection of H across the midpoint of be Y. Then X and Y both lie on the circumcircle of. oreover, Y is a diameter of the circumcircle. H X Y Proof. Trivial. ngle chasing. (i) Prove the existence of the nine-point circle. (Given a triangle, the nine-point circle is the circle that passes through the three midpoints of sides, the three feet of altitudes, and the three midpoints between the orthocenter and the vertices). (ii) Let be a triangle, and P a point on its circumcircle. Show that the reflections of P across the three sides of lie on a lie that passes through the orthocenter of. (iii) (O Shortlist 2005) Let be an acute-angled triangle with, let H be its orthocentre and the midpoint of. Points on and on are such that = and, H, are collinear. Prove that H is orthogonal to the common chord of the circumcircles of triangles and. (iv) (US TST 2005) Let be an acute triangle, and let O and H be its circumcenter and orthocenter, respectively. or 1 i 3, points P i and Q i lie on lines O i and i+1 i+2 (where i+3 = i ), respectively, such that OP i HQ i is a parallelogram. Prove that OQ 1 OP 1 + OQ 2 OP 2 + OQ 3 OP 3 3. (v) (hina TST quizzes 2006) Let ω be the circumcircle of triangle, and let P be a point inside the triangle. Rays P,P,P meet ω at 1, 1, 1, respectively. Let 2, 2, 2 be the images of 1, 1, 1 under reflection about the midpoints of,,, respectively. Show that the orthocenter of lies on the circumcircle of O and H are isogonal conjugates. Let be a triangle, with circumcenter O, orthocenter H, and incenter. Then is the angle bisector of HO. Proof. Trivial. 9

10 (i) (rux) Points O and H are the circumcenter and orthocenter of acute triangle, respectively. The perpendicular bisector of segment H meets sides and at and, respectively. Prove that O = O. (ii) Show that H = O if and only if one of,, is