Let $ ABC$ be an acute-angled triangle. The lines $ L_{A}$, $ L_{B}$ and $ L_{C}$ are constructed through the vertices $ A$, $ B$ and $ C$ respectively according the following prescription: Let $ H$ be the foot of the altitude drawn from the vertex $ A$ to the side $ BC$; let $ S_{A}$ be the circle with diameter $ AH$; let $ S_{A}$ meet the sides $ AB$ and $ AC$ at $ M$ and $ N$ respectively, where $ M$ and $ N$ are distinct from $ A$; then let $ L_{A}$ be the line through $ A$ perpendicular to $ MN$. The lines $ L_{B}$ and $ L_{C}$ are constructed similarly. Prove that the lines $ L_{A}$, $ L_{B}$ and $ L_{C}$ are concurrent.
Problem
Source: IMO ShortList 1988, Problem 15, Iceland 1, Problem 34 of ILL
Tags: geometry, circumcircle, Triangle, concurrency, IMO Shortlist
Arne
22.10.2005 19:05
I posted a solution to a very similar problem a few days ago. The lines $L_{A}$, $L_{B}$ and $L_{C}$ are the isogonal conjugate lines of the altitudes of the triangle. Hence they concur at the isogonal conjugate point of the orthocentre, which is the circumcentre.
darij grinberg
24.09.2006 13:42
For the sake of completeness, an angle chase solution: Let O be the circumcenter of triangle ABC. We will prove that the lines $L_{A}$, $L_{B}$, $L_{C}$ are concurrent at the point O. We work with directed angles modulo 180°. Since the points A, H, M, N all lie on the circle $S_{A}$, we have < AMN = < AHN. Since the point N lies on the circle $S_{A}$ with diameter AH, we have < ANH = 90°, so that, in the right-angled triangle ANH, we have < AHN = 90° - < NAH. But since < AHC = 90°, in the right-angled triangle AHC we have < HCA = 90° - < CAH. Thus, < AMN = < AHN = 90° - < NAH = 90° - < CAH = < HCA = < BCA. On the other hand, the point O is the circumcenter of triangle ABC, thus the center of a circle through the points A, B, C; hence, by the central angle theorem, < OAB = 90° - < BCA. Thus, < (AO; MN) = < (AO; AB) + < (AB; MN) = < OAB + < AMN = (90° - < BCA) + < BCA = 90°, so that $AO\perp MN$. Thus, the line $L_{A}$, being defined as the perpendicular to MN through A, must coincide with the line AO, and therefore pass through the point O. Similarly, the lines $L_{B}$ and $L_{C}$ pass through O, and therefore the lines $L_{A}$, $L_{B}$, $L_{C}$ are concurrent at the point O. Problem solved. darij
Pascal96
14.04.2011 14:45
My solution is the same as Arne's. If we use the properties of isogonal conjugates like that in an olympiad, do we have to prove or justify them?
Jzhang21
12.12.2018 04:59
We claim that the desired intersection point is $O$, the circumcenter of $\triangle ABC$. Let $X$ be the feet from $A$ to $MN$. Note that $\angle HAC=90^{\circ}-\angle C.$ Hence, $\angle AMX=\angle HNA=90^{\circ}-\angle HAN=\angle C.$ Therefore, $$\angle MAX=90^{\circ}-\angle C=\angle BAO$$so $O$ lies on $AX$, which is $L_A.$ We can similarly show that $O$ lies on $L_B$ and $L_C.$ $\blacksquare$