Point in a triangle, with angle constraints

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Point in a triangle, with angle constraints

G H J

Reveal topic tags

geometric transformation

L

G H BBookmark kLocked kLocked NReply

Source: Tuymaada 2012, Problem 3, Day 1, Seniors

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

15142 posts

#1 h Jul 21, 2012, 9:38 PM • 6 Y

Y by Adventure10, Mango247, and 4 other users

Point $P$ is taken in the interior of the triangle $ABC$ , so that
$\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C).$
Let $L$ be the foot of the angle bisector of $\angle B$ . The line $PL$ meets the circumcircle of $\triangle APC$ at point $Q$ . Prove that $QB$ is the angle bisector of $\angle AQC$ .

Proposed by S. Berlov

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

71 posts

#2 h Jul 23, 2012, 3:06 PM • 2 Y

Y by Adventure10, Mango247

Dear friend from the beautiful Montevideo, one quick thought:

♦ All in this figure is symmetric, with symmetry axis the line (e) which is defined by the bisector of the angle $B$ .
$L$ is the polar of point $B$ with respect of the circle (g).
Hence, this point is also the meeting point of the diagonals of the isosceles trapezium $TSQ_1Q_2$ , with $S$ the midpoint of the arc $CSA$ and $T$ the midpoint of the arc $A_1TC_1$ .
We have here:
$\angle CQ_1 S = \angle SQ_1 A,\;\;A_1 T = TC_1 \Rightarrow \angle TCB$ $= \angle TAB = \frac{{\angle A + \angle C}} {4}\; \Rightarrow T \equiv P,\;\;Q_1 \equiv Q,$
because $\angle ATC = 90^ \circ + \frac{{\angle B}} {2}$

Attachments:

art.pdf (237kb)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

212 posts

Lawasu

#3 h Jul 24, 2012, 10:24 AM • 2 Y

Y by Adventure10, Mango247

WLOG $AB>BC$ . By the hypothesis we get $\widehat{PAC}+\widehat{PCA}=\dfrac{1}{2}\cdot (\widehat{A}+\widehat{C})$ , so $\widehat{APC}=90^\circ +\widehat{B}/2$ .
Therefore $AIPC$ is cyclic where $I$ is the incenter of $\triangle ABC$ . We know that $[BL$ passes through the center $W$ of circle $(AIC)$ .
Hence, the reflection $M$ of $A$ about $BL$ lies on the circle $(AIC)$ and on the line $BC$ .
The reflection $N$ of $C$ about $BL$ also lies on the circle $(AIC)$ and on the line $AB$ .
Let $R$ be the reflection of $P$ about line $BL$ . By the symmetry we get $\widehat{RMC}=\widehat{PAN}=\widehat{PCB}$ .
Therefore $PC\parallel RM$ , so $arc\ RP=arc\ CM$ and it follows that $\overarc{AR}=\overarc{RC}$ since $AM\parallel CN\parallel PR$ .
Hence $R$ is the midpoint of arc $AC$ and $P$ is the midpoint of the arc $MN$ . Let $I_b$ be the $B$ -excenter of $\triangle ABC$ .
The division $(I,I_n,B,L)$ is harmonic, so $L$ lies on the polar of $B$ with respect to the circle $(AIC)$ .
Now take $R_1\in BR\cap (AIC),\ R_1\neq R$ and $P_1\in BP\cap (AIC),\ P_1\neq P$ . Take $K\in RP_1\cap PR_1$ .
Obviously $K$ lies on $BL$ and it's well known that $K$ lies on the polar of $RR_1\cap PP_1$ with respect to the circle $(AIC)$ .
Hence $K=L$ and it follows that $R_1=Q$ since $Q\in PL\cap (AIC)$ . Now, obviously $[QB$ is the bisector of $\widehat{AQC}$ since $B,Q,R$ are collinear.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

339 posts

leader

#4 h Jul 24, 2012, 11:30 AM • 1 Y

Y by Adventure10

Find a point $X$ in the plane such that $\angle ABX=\angle CBP$ (where the line $BX$ cuts segment $AC$ ) and $\angle AXB=\angle PCB$ , and prove $X\equiv Q$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

111 posts

#5 h Apr 26, 2014, 4:37 PM • 2 Y

Y by Adventure10, Mango247

o(center of circumcircle of triangle APL) is midpoint of AC in circumcircle of ABC.
let angle bisector of B meet AC at T. OT.TB=AT.CT=PT.TQ ----> OPBQ is cycle
now it is easy by play with angle.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

314 posts

#6 h Aug 28, 2014, 2:22 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the midpoint of the arc $AC$ .Now,it is easy to obtain $LA*LC=LB*LM=LP*LQ$ ,so we obtain $BPMQ$ is a cyclic and the rest is easy angle chasing.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1079 posts

#7 h Oct 1, 2021, 4:00 PM

Y by

Claim: Let $C'$ be the reflection of $C$ over the angle bisector of $\angle ABC$ . Then, $BPCC'$ is cyclic.
Proof.Indeed, $\begin{align*} \measuredangle PCC'&=\measuredangle ACC'+\measuredangle PCA=\frac{\measuredangle B+\measuredangle C}{2}-\frac{\measuredangle B+\measuredangle C}{4}\\&=\frac{\measuredangle B+\measuredangle C}{4}=\measuredangle PBA=\measuredangle PBC'.\,\square \end{align*}$
Let $Q'$ be image of $P$ of inversion at $B$ with radius $\sqrt{ac}$ followed by the reflection over the angle bisector of $\angle ABC$ . Thus, $Q'B$ is the angle bisector of $\angle CQ'A$ and $APCQ'$ is cyclic quadrilateral. Now, all we need is to show that $P,L,Q'$ are collinear.
By ratio lemma, $\begin{align*} \frac{AL}{LC}=\frac{AB}{BC}=\frac{AB}{BC}\cdot \frac{\sin{\angle ABP}}{\sin{\angle PBC}}\cdot \frac{\sin{\angle ABQ'}}{\sin{\angle Q'BC}}=\frac{AP}{PC}\cdot \frac{AQ'}{Q'C}, \end{align*}$ we are done. $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a