right angle wanted, AP: PB = CQ: QA = 2 inside an equilateral

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

right angle wanted, AP: PB = CQ: QA = 2 inside an equilateral

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: All-Siberian Open School Olympiad 2014-15 10.3

The post below has been deleted. Click to close.

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

30629 posts

#1 h Jul 4, 2020, 10:48 AM • 1 Y

Y by Mango247

In an equilateral triangle $ABC$ on the sides $AB$ and $AC$ taken points $P$ and $Q$ respectively, such that $AP: PB = CQ: QA = 2$ . Let $O$ be the intersection point of the segments $CP$ and $BQ$ , prove that the angle $AOC$ is right ..

Z K Y

The post below has been deleted. Click to close.

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

27 posts

#2 h Jul 4, 2020, 4:33 PM • 2 Y

Y by Mango247, Mango247

Since $ABC$ is equilateral, $BP=AQ$ . Because we have $BP=AQ$ , $BC=AB$ , and $\angle PBC=\angle QAB$ , by SAS, $PBC \cong QAB$ . Hence $\angle BPC = \angle AQB$ and $\angle OPA = 180^{\circ} -\angle OPB = 180^{\circ}-\angle OQA$ , which means $AQOP$ is cyclic. Now in triangle $AQP$ , let $AQ=x$ . Then $AP=AB-BP=3x-x=2x$ . Furthermore, we have by cosine law, $PQ^2=AQ^2+AP^2-2\cdot AQ \cdot AP \cos(\angle PAQ)=x^2+(2x)^2-2(x)(2x)\cos(60^{\circ})=5x^2-4x^2(\frac12)=3x^2$ Now, notice $PQ^2+AQ^2=AP^2$ , so by Pythagoras', $\angle AQP$ is a right angle (remarkably, $AP$ is the diameter of the circumcircle). Thus, by cyclic quad., $\angle AOP = 90^{\circ}$ , and the supplementary angle, $\angle AOC$ , is also right. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

258 posts

#3 h Jul 4, 2020, 7:29 PM

Y by

From the fact that triangles BPC=AQB we have angles BPC=AQB then AQOP is cyclic so angle COQ=BAC=60
Let R the midpoint of AP and S the midpoint of CQ then ASP is an equaliteral triangle with RQ as its A-midline then RQ=1/2*PS and AR=PR=1/2*PS so RQ=AR=PR then angle AQP=90 then angle APQ=30 but because AQOP os cyclic we have angle AOQ=APQ=30
Now angle COA=COQ+AOQ=60+30=90 as desired

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a