SOME PROPERTIES OF INTERSECTION POINTS OF EULER LINE AND ORTHOTRIANGLE

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1 SOME PROPERTIES OF INTERSETION POINTS OF EULER LINE ND ORTHOTRINGLE DNYLO KHILKO bstract. We consider the points where the Euler line of a given triangle meets the sides of its orthotriangle, i.e. the triangle whose vertices are feet of the altitudes of. In this note we study properties of these points and how they relate to the known objects. notable construction occured in [2, Problem 3] and [1, Problem G6]. problem inspired by this construction initiated the research which we present in this paper. While solving this problem, we discovered several facts about intersection points of the Euler line and the sides of the orthotriangle. Further investigation of these points resulted in facts which we find intersting on their own and decided to share them. The following notation will be used. Let be an acute triangle. Its altitudes H,, intersect at the orthocenter H. Denote the midpoints of the sides,, by, M,, respectively, and the circumcenter of by O. Let X be the foot of the perpendicular from to. Define the points X, X analogously. Let us remind reader some classical facts first. The points H,,, M,, lie on a circle (the nine-point circle) centered at O 9 which is the midpoint of OH. The lines X, X, X meet at the point O. The next lemma can be used to prove various facts including IMO and IMOSL2012 G6. Lemma 1. The circumcircles of the triangles M X X, X X, X X intersect at the point O. Proof. onsider the circle ω with diameter OH. Since X, X, X meet at O, we have H X O = H X O = H M O = 90. Then the points H, M, X, X lie on ω. Similarly we have that the circumcircles of the triangles M X X, X X, X X intersect at the point O. One might wonder, whether a similar statement holds for the triangles M X, M X, X. The following theorem provides the answer. Theorem 1. The circumcircles of the triangles M X, M X, X have a common point which belongs to the Euler line. efore proving Theorem 1, we establish an auxiliary result. object of the proof, which is the main object of this exposition. It introduces the key Proposition 1. Let OH intersects the lines, H, H at the points K, K, K. Then the points,, X, K are cyclic. The same holds for the fours of points M,, X, K ;, M, X, K. 35

2 36 DNYLO KHILKO We need the following lemma which was proposed on the ll-russian Mathematical Olympiad [3, , District round, Grade 11, Problem 4] Lemma 2. Let the lines and meet at the point T. Then T OH. ω 1 ω 2 N O T H Fig. 1. Proof. Denote by ω 1 the circumcircle of (see Fig. 1)). Let OH intersect ω 1 again at the point N. Then H = H = NH = 90. We have that 90 = NO = O = O. Hence N lies on the circumcircle of, denoted by ω 2. onsider the circles ω 1, ω 2 and the nine-point circle of. The line N is the radical axis of ω 1 and ω 2. The line is the radical axis of ω 1 and the ninepoint circle. Finally, the line is the radical axis of ω 2 and the nine-point circle. This implies that the line N passes through T. Then T OH. ω 1 ω 2 N O T K X H Fig. 2. Proof of Proposition 1. We will show that T T = T X T K (see Fig. 2). y Lemma 2, NH = NK = X K = 90. Then, N, X, K are concyclic. We obtain the following equation T N T = T X T K.

3 SOME PROPERTIES OF INTERSETION POINTS OF EULER LINE ND ORTHOTRINGLE 37 lso we have T N T = T T Hence T T = T X T K. Thus the points,, X, K are concyclic. Now we are ready to prove Theorem 1. O S K X H X K H Fig. 3. Proof of Theorem 1. Let the circumcircle of the quadrilateral X K M meets OH again at S (see Fig. 3). It suffices to show that the point S belongs to the circumcircle of X K, a similar statement for M X K will follow. We will work with oriented angles between lines. Denote by (l, m) the angle of the counterclockwise rotation which maps a line l to one parallel to a line m. See more in [4]. We have ( X, X K ) = ( S, SK ) = ( S, SK ). From Lemma 2 we obtain that ( X, X K ) = ( X, X ) = ( X, X ) = ( X, X K ). Hence we conclude that and the proof is completed. ( S, SK ) = ( X, X K ), Remark 1. It is possible to prove the first part of Theorem 1 about three circles by angle chasing using Lemma1, however, this way does not imply that the intersection point of the circles lies on the line OH. Having proved Theorem 1, we establish further properties of the points K, K, K. Theorem 2. The circumcircles of the triangles K H O 9, K O 9 and K O 9 have another common point different from O 9.

4 38 DNYLO KHILKO Firstly, we remind that in Lemma 2 we have defined the point T as the common point of and. Define the points T and T analogously. Proposition 2. The points K, H, T and O 9 are cyclic. 1 N O T O9 K X H H Fig. 4. Proof. In the proof of Proposition 1 we have obtained that the points, N, X and K are cyclic. Therefore, (X, T ) = (X, N) = (X K, K N) = (T K, K O 9 ). We claim that (X, T ) = (T H, H O 9 ) (see Fig. 4). Indeed, the points and H are symmetric with respect to M. Then the nine-point circle and the circumcircle of M are symmetric with respect to M. Hence O 9 is symmetric to the point 1 which is the center of the circumcircle M. s O belongs to this circle and NO = 90 we have that 1 lies on O i. e. X. So we have (X, T ) = (T H, H O 9 ). Then (T H, H O 9 ) = (X, N) = (T K, K O 9 ), and we are done. onsider the points T, T, T. Lemma 3. The points T, T, are collinear.

5 SOME PROPERTIES OF INTERSETION POINTS OF EULER LINE ND ORTHOTRINGLE 39 T T Q L H Fig. 5. Proof. We need some additional notation, which will be used only in this proof. Denote by Q the intersection point of and and by L the intersection point of Q and (see Fig. 5). Let us apply Desargues theorem for the triangles H Q and the one formed by the lines M,, (this triangle has one vertex at infinity). Then the following statements are equivalent:, Q and the line parallel to passing through H are concurrent and the intersection points of Q and, H and M, QH and are collinear. Notice that H meets M at T, Q meets at T and QH meets at. So in order to prove that T, T, are collinear we will prove the first statement obtained by Desargues theorem. It is sufficient to prove that LH. y Menelaus theorem Then It is a well-known fact that Hence and LH. Q Q L L = 1. L L = Q Q. Q Q = H H. L L = H H, Remark 2. similar fact to Lemma 3 will hold if one replace the points H,, by some points 1, 1, 1 which lie on the respective sides of and 1, 1, 1 are concurrent. The next fact describes other properties of T, T, T. Lemma 4. The circumcircles of the triangles T T, T T H and T T have a common point P. Proof of Lemma 4. The statement follows from Miquel s theorem applied to the triangle H and the points T, T, T (see Fig. 6).

6 40 DNYLO KHILKO T T P T H Fig. 6. M T T T P S O 9 K H M Fig. 7.

7 SOME PROPERTIES OF INTERSETION POINTS OF EULER LINE ND ORTHOTRINGLE 41 Now we claim that the point P lies on the circumcircle of the triangle K H O 9. This fact combined with that for K O 9 and K O 9 is equivalent to Theorem 2. Proof of Theorem 2. In order to prove that the circumcircle of O 9 K H passes through P, we will show that (T P, P H ) = (T O 9, O 9 H ) (See Fig. 7). Firstly, we will prove that O 9 T T T. Let and intersect at U. It is a well-known fact that U is the polar line of T with respect to the nine-point circle. pplying Pascal s theorem on the hexagon M H we obtain that T, T, U are collinear. Hence T T is the polar line of T, consequently, O 9 T T T. Let S be the foot of the perpendicular from O 9 to H. Then (T O 9, O 9 S) = (T T, T S). lso we have (SO 9, O 9 H ) = (, H ). So (T O 9, O 9 H ) = (T O 9, O 9 S) + (SO 9, O 9 H ) = = (T T, T S) + (, H ) = = (T T, T H ) + (T, T ) = = (T P, P H ) + (T P, P T ) = (T P, P H ) and we are done. Theorem 3. T T, and K H are concurrent. Proposition 3. K H is tangent to the circumcircle of T T H at H. Proof. We have (K H, H P ) = (K T, T P ) = = ( T, T P ) = ( T, T ) = (H T, T P ). T T Q P T O 9 K Fig. 8. H M

8 42 DNYLO KHILKO Proof of Theorem 3. Let T T intersect K H at Q (see Fig. 8). y Proposition 3 QH is tangent to the circumcircle of the triangle T H T. lso H bisects T H T. Hence Q = QH since QH = T T H + H T = T H Q + T H = QH. So Q belongs to the perpendicular bisector of H, which is obviously M. The author is grateful to Vladyslav Vovchenko for a short proof of the Lemma 3 and to Georgiy Shevchenko for careful reading of the manuscript. References [1] 53-rd International Mathematical Olympiad, Shortlisted Problems with Solutions url: [2] 54-th International Mathematical Olympiad url: org/problems.aspx. [3] N. K. gakhanov, I. I. ogdanov, P.. Kozhevnikov, O. K. Podlipsky, and D.. Tereshin. ll-russian mathematical olympiad M.:MME, [4] V. V. Prasolov. Problems in plane geometry. Russian. M.:MME, Taras Shevchenko National University of Kyiv address: dkhilko@ukr.net