Midpoints of chords of an excircle through the circumcircle

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

Midpoints of chords of an excircle through the circumcircle

G H J

Reveal topic tags

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: XVIII Tuymaada Mathematical Olympiad (2011), Junior Level

The post below has been deleted. Click to close.

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

316 posts

Shu

#1 h Jul 29, 2011, 6:08 PM • 2 Y

Y by Adventure10, Mango247

An excircle of triangle $ABC$ touches the side $AB$ at $P$ and the extensions of sides $AC$ and $BC$ at $Q$ and $R$ , respectively. Prove that if the midpoint of $PQ$ lies on the circumcircle of $ABC$ , then the midpoint of $PR$ also lies on that circumcircle.

Z K Y

The post below has been deleted. Click to close.

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

58 posts

Spats

#2 h Jul 31, 2011, 10:30 AM • 1 Y

Y by Adventure10

Denote by $S$ the midpoint of $PQ$ and $T$ the midpoint of $PR$ . Because $AP=AQ$ and $BP=BR$ we find that $[AS$ is the bisector of $\angle{PAQ}$ and $[BT$ is the bisector of $\angle {PBR}$ .
$m(\angle STP)=m(\angle QRP)=m(\angle ARQ)-m(\angle BRP)= 90^{\circ}-\frac{m(C)}{2}-\frac{m(B)}{2}=\frac{m(A)}{2} \Rightarrow m(\angle STB)=90^{\circ}+\frac{m(A)}{2}$
$m(\angle SAB)= 90^{\circ}-\frac{m(A)}{2}$ .

We obtain that $m(\angle SAB)+m(\angle STB)=180^{\circ}$ , so the quadrilateral $SABT$ is cyclic. q.e.d

Z K Y

The post below has been deleted. Click to close.

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

4384 posts

#3 h Jul 31, 2011, 11:12 AM • 2 Y

Y by Adventure10, Mango247

Following the same idea, i.e. to prove that $ASTB$ is cyclic looks much easier this way:
$AS$ and $BT$ concur at $I_a$ , the $A$ -excenter, so $I_aQ\perp AC$ and $I_aR\perp BC$ ; also $I_aQ=I_aR$ ; consequently, $I_aQ^2=I_aS\cdot I_aA=I_aT\cdot I_aB=I_aR^2$ and, indeed, $ASTB$ is cyclic.

Best regards,
sunken rock

Z K Y

The post below has been deleted. Click to close.

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

125 posts

#4 h Nov 12, 2020, 10:53 AM

Y by

We consider the $A-$ excircle. let $I_A$ be the $A-$ excenter, $P$ , $Q$ , $R$ be the touchpoints of the $A-$ excircle with $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ , respectively. Let $M$ , $N$ be the midpoints of $\overline{PR}$ , $\overline{PQ}$ , respectively. Then obviously $B$ , $M$ , $I_A$ are collinear and $\overline{I_AP} \perp \overline{BC}$ , $\overline{PM} \perp \overline{BI_A} \implies I_AM \cdot I_AB=IP^2$ . Similarly $I_AN \cdot I_AC=IP^2$ . Thus $MNBC$ is cyclic and $ABCM$ is cyclic $\iff$ $ABCN$ is cyclic, as desired.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a