Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \neq CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \neq C$ on the line $CL$, we have $\angle XAC \neq \angle XBC$. Also show that for $Y \neq C$ on the line $CH$ we have $\angle YAC \neq \angle YBC$. Bulgaria
Problem
Source: JBMO 2001, Problem 2
Tags: geometry, circumcircle, symmetry, trigonometry, perpendicular bisector, angle bisector, power of a point
yetti
02.11.2005 08:59
(1) Let $O_A, O_B$ be the circumcenters and $R_A, R_B$ the circumradii of the triangles $\triangle XAC, \triangle XBC$ with the common side XC. It is sufficient to show that the circumradii $R_A \neq R_B$ are not equal. The perpendicular bisectors of the sides CA, CB meet at the circumcenter O of the right angle triangle $\triangle ABC$, i.e., at the midpoint of the hypotenuse AB. The angle $\angle OO_AO_B = 90^\circ$ is right, because $OO_A, OO_B$ are the perpendicular bisectors of the triangle sides CA, CB and $CA \perp CB$. The angle $\angle OO_AO_B = OO_BO_A = 45^\circ$, because the $O_AO_B$ is the perpendicular bisector of the angle bisector segment CX, which forms the angle $45^\circ$ with both CA, CB and $OO_A \perp CA, OO_B \perp CB$. Thus the triangle $\triangle OO_AO_B$ is isosceles and its symmetry axis, the perpendicular bisector of the segment $O_AO_B$ is parallel to the angle bisector CL, identical with the radical axis of the circles $(O_A), (O_B)$. Therefore, these 2 circles do not intersect on the perpendicular bisector of the segment $O_AO_B$, which implies that their radii are different, $R_A \neq R_B$. (2) Let $O_1, O_2$ be the circumcenters and $R_1, R_2$ the circumradii of the triangles $\triangle YAC, \triangle YBC$ with the common side YC. It is sufficient to show that the circumradii $R_1 \neq R_2$ are not equal. The perpendicular bisectors of the sides CA, CB meet at the circumcenter O of the right angle triangle $\triangle ABC$, i.e., at the midpoint of the hypotenuse AB. The angle $\angle OO_1O_2 = 90^\circ$ is right, because $OO_1, OO_2$ are the perpendicular bisectors of the triangle sides CA, CB and $CA \perp CB$. The angles $\angle OO_1O_2 = \angle CBA$ are equal and the angles $\angle OO_2O_1 = \angle CAB$ are equal, because the the perpendicular bisector $O_1O_2$ of the altitude segment CY is parallel to the hypotenuse AB, both being perpendicular to the altitude CH. Thus the triangles $\triangle O_2O_1O \sim \triangle ABC$ are centrally similar, having their corresponding sides parallel. The lines connecting the corresponding vertices C, O pass through their similarity center. Since O is also the midpoint od AB, the line CO intersects the segment $O_1O_2$ at its midpoint M. Since CO is not parallel to the altitude CH and these 2 lines meet at C, the midpoint M does not lie on the altitude CH. The perpendicular bisector of the segment $O_1O_2$ passing through M is parallel to the altitude CH, identical with the radical axis of the circles $(O_1), (O_2)$, both being perpendicular to $AB \parallel O_1O_2$. Therefore, these 2 circles do not intersect on the perpendicular bisector of the segment $O_1O_2$, which implies that their radii are different, $R_1 \neq R_2$.
Attachments:
NuncChaos
05.12.2010 05:47
Part 1: If $\angle XAC$ = $\angle XBC$, then since we are given that $\angle ACX$ = $\angle BCX$, we have $\angle BXC$ = $\angle AXC$, which implies congruence by ASA. Thus, $\frac{XC}{BC}=\frac{XC}{AC} \longrightarrow AC=BC$, a contradiction. Part 2: Since $\bigtriangleup$ CAH and $\bigtriangleup$ BCH are similar, we have $\frac{AH}{AC}=\frac{BH}{BC} \qquad a (1)$ By law of sines, we have $\frac{sin CAY}{CY} = \frac{sin AYC}{AC}$ But, $sin(AYC) = sin (AYH) = \frac{AH}{AY}$, so plugging in yields $\frac{sin(CAY)}{CY}=\frac{AH}{(AY)(AC) } \qquad a (2)$. Similarly, we have $\frac{sin(CBY)}{CY}=\frac{BH}{(BY)(BC) } \qquad a(3)$. But if $\angle CBY$ = $\angle CAY$, $sin CBY = sin CAY$, so from (2) and (3) we now have: $\frac{AH}{(AY)(AC)} = \frac{BH}{(BY)(BC)}$ Plugging in (1) yields $\frac{1}{AY}=\frac{1}{BY} \longrightarrow AY=BY$ Thus $\bigtriangleup$ AYB is isosceles. This forces $\angle XAB = \angle XBA \longrightarrow \angle BAC = \angle ABC \longrightarrow BC=AC$, a contradiction.
epsilon07
29.09.2011 10:05
Part 2: Since $\bigtriangleup$ CAH and $\bigtriangleup$ BCH are similar, we have $\frac{AH}{AC}=\frac{HB}{BC} \qquad a (1)$ Similarity you used is wrong.. it must be $\frac{AH}{AC}=\frac{HC}{BC} \qquad a (1)$
butter67
20.12.2013 02:04
All we need to prove is the contrapositive. Proceed as NuncChaos did.
All we need to prove is the contrapositive. Let $b=CY$ and $a=\angle{CAY}=\angle{YBC}$. We have $\frac{b}{\sin{a}}$ as twice the circumradius of $\triangle{ACH}$ and $\triangle{BCH}$. But twice the circumradius is $AC$ and $BC$! Contradiction.
efang
28.07.2015 23:36
I did part 1 the same as NuncChaos did
Assume there did exist a point Y on CH such that $\angle{YAC} = \angle{YBC}$
Let $A'$ be on AC and $B'$ on BC such that $YA' \perp AC$ and $YB' \perp BC$. Clearly $HBB'Y$ and $HAA'Y$ are cyclic. This means that $\angle{YBC} = \angle{YHB'}$ and $\angle{YAC} = \angle{YHA'}$
Notice that $A'B'$ is bisected by $YC$ because they are 2 diagonals of the same rectangle. Thus $HY$, which is the angle bisector of $\triangle{HA'B'}$, also bisects $A'B'$. Thus by the angle bisector theorem we must have $HA' = HB'$ so $YHA' \equiv YHB'$ by SAS.
This means that $YA' = YB'$, thus implying that $Y$ lies on the angle bisector. But since $AB \neq AC$ the triangle is not isoseles and the height and angle bisector only meet at C. But since $Y \neq C$ we arrive at a contradiction. Thus no point Y may exist.