2001 JBMO Problems/Problem 2

Revision as of 00:27, 14 August 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 2 (part 2 credit to efang) -- one easier and one harder)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \ne CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \ne C$ on the line $CL,$ we have $\angle XAC \ne \angle XBC$. Also show that for $Y \ne C$ on the line $CH$ we have $\angle YAC \ne \angle YBC$.

Solution

[asy] draw((0,0)--(30,0)--(0,40)--(0,0)); dot((0,0)); label("C",(0,0),SW); dot((0,40)); label("A",(0,40),NW); dot((30,0)); label("B",(30,0),SE);  draw((0,0)--(120/7,120/7)); dot((120/7,120/7)); label("L",(120/7,120/7),NE);  dot((10,10)); label("X",(10,10),S); draw((0,40)--(10,10)--(30,0));  draw(anglemark((0,0),(0,40),(10,10),150)); draw(anglemark((10,10),(30,0),(0,0),150)); [/asy]

Assume that $\angle XAC = \angle XBC$. Since $CL$ is an angle bisector of $\angle ACB,$ $\angle ACX = \angle BCX.$ Thus, by AAS Congruency, $\triangle ACX \cong \triangle BCX,$ which results in $AC = BC.$ But $AC \ne BC,$ so by proof by contradiction, $\angle XAC \ne \angle XBC.$

[asy] draw((0,0)--(30,0)--(0,40)--(0,0)); dot((0,0)); label("C",(0,0),SW); dot((0,40)); label("A",(0,40),NW); dot((30,0)); label("B",(30,0),SE);  draw((0,0)--(96/5,72/5)); dot((96/5,72/5)); label("H",(96/5,72/5),NE);  dot((12,9)); label("Y",(12,9),NW); draw((0,40)--(12,9)--(30,0)); draw((0,9)--(12,9)--(12,0),dotted);  dot((0,9)); label("$A'$",(0,9),W); dot((12,0)); label("$B'$",(12,0),S);  draw(anglemark((0,0),(0,40),(12,9),150)); draw(anglemark((0,0),(0,40),(12,9),200)); draw(anglemark((12,9),(30,0),(0,0),150)); draw(anglemark((12,9),(30,0),(0,0),200));  [/asy]

Assume that $\angle YAC = \angle YBC$. Draw points $A'$ and $B'$ on $AC$ and $BC$ respectively such that $AY'BC$ is a rectangle. That means $B'BHY$ and $AA'YH$ are cyclic quadrilaterals, which means that $\angle A'HY = \angle B'HY.$


Because $A'B'$ is bisected by $HC$, by the Angle Bisector Theorem, we have $HA' = HB',$ so by SAS Congruency, we have $\triangle HA'Y \cong \triangle HB'Y,$ making $A'Y = YB'$ and $A'YCB'$ a square. This also means that $\angle A'CY = \angle B'CY,$ so $\triangle ACY \cong \triangle BCY$ by AAS Congruency, making $AC = BC.$ However, it is given that $AC \ne BC,$ so $\angle YAC \ne \angle YBC$ by proof by contradiction.

See Also

2001 JBMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4
All JBMO Problems and Solutions
Invalid username
Login to AoPS