prove that a chord is tangent to the incircle

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

w

26 users

TOPICS

No tags match your search

M

No tags match your search

M

prove that a chord is tangent to the incircle

G H J

Reveal topic tags

incenter config

L

G H BBookmark kLocked kLocked NReply

Source: SORY 2019 P6

The post below has been deleted. Click to close.

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

168 posts

#1 h Mar 23, 2025, 9:01 PM

Y by

Let $ABC$ be a triangle with incenter $I$ and intouch triangle $DEF$ . Let $P$ be the foot of the perpendicular from $D$ onto $EF$ . Assume that $BP$ , $CP$ intersect the sides $AC$ , $AB$ in $Y,Z$ respectively. Finally, let the rays $IP$ , $YZ$ meet the circumcircle of $\triangle ABC$ in $R$ , $X$ respectively. Prove that the tangent from $X$ to the incircle and the line $RD$ meet on the circumcircle of $\triangle ABC$ .

Proposed by Aditya Khurmi

This post has been edited 1 time. Last edited by ihategeo_1969, Mar 23, 2025, 9:08 PM

Z K Y

The post below has been deleted. Click to close.

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

168 posts

#2 h Mar 23, 2025, 9:07 PM • 1 Y

Y by cursed_tangent1434

Why so cursed?
Solution

Z K Y

The post below has been deleted. Click to close.

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

cursed_tangent1434

557 posts

cursed_tangent1434

#3 h Mar 24, 2025, 1:54 AM • 1 Y

Y by ihategeo_1969

Not cursed really, pretty standard. Let $M$ be the minor $BC-$ arc midpoint in $(ABC)$ . It is well known that points $R$ , $D$ and $M$ are collinear, so we simply need to show that the tangent from $X$ to the incircle passes through $M$ which we shall do as follows.

Let $X$ and $W$ be the intersections of the tangents from $M$ to the incircle with $(ABC)$ . From this solution to LMAO 2021/5 we know that $X$ and $W$ are the the intersections of the circle with center $A$ and radius $AI$ with $(ABC)$ and that $XW$ is a tangent to the incircle. Now, let $Z$ and $Y$ be the intersections of $XW$ with sides $AC$ and $AB$ respectively. We can make the following observation.

Claim : Quadrilateral $BCYZ$ is cyclic.

Proof : Consider the inversion centered at $A$ with radius $AI$ . Points $X$ and $W$ are clearly fixed under this transformation so $Z = XW \cap AB$ will map to $(AXW) \cap AB$ which is just $B$ . Similarly $Y$ maps to $C$ so points $B$ , $C$ , $Y$ and $Z$ must be concyclic.

With the claim in hand, we are not left with an angle chase. Let $K$ denote the tangency point of $XW$ with the incircle. Then,
$\begin{align*} \measuredangle (KD,EF) &= \measuredangle EDK + \measuredangle FED \\ &= \measuredangle YEK + \measuredangle FED \\ &= \measuredangle CBI + \measuredangle FDB\\ &= 90^\circ \end{align*}$ which implies that $DK \perp EF$ . Thus, points $D$ , $P$ and $K$ are collinear. Now, by Brianchon's Theorem on tangential quadrilateral $BCYZ$ we have that lines $BY$ , $CZ$ , $DK$ and $EF$ concur. The latter two intersect at $P$ which implies that $Y = BP \cap AC$ and $Z=CP \cap AB$ which coincide with the definitions in the problem. Thus, we are done.

Z K Y

The post below has been deleted. Click to close.

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

667 posts

#4 h Yesterday at 9:23 AM

Y by

Let $S$ be $A-$ sharky devil (won't use $R$ for that), $M$ be the midpoint of arc $BC$ , let $MM_1,MM_2$ be tangent to the incircle and $M_1,M_2\in (ABC)$ . Let $AS\cap BC=Q,QH$ be tangent to the incircle where $H$ lies on it and $QH$ meet $AB,AC$ at $K,L$ . Note that $S,D,M$ are collinear.

DDIT gives $(\overline{MA},\overline{MD}),(\overline{MB},\overline{MC}),(\overline{MM_1},\overline{MM_2})$ is an involution. Project this onto $(ABC)$ to get $AS,M_1M_2,BC$ concur thus, $Q,M_1,M_2$ are collinear. Note that $DH\perp QI\perp AI\perp EF$ hence $DH\perp EF$ which implies $D,P,H$ are collinear. Since $EF,DH$ are polars of $A,Q$ we see that $DF\cap EH,DE\cap HF$ lie on $AQ$ . Also $P$ is on the polar of $A$ so by La Hire $A$ is on the polar of $P$ . Also the polar of $P$ is perpendicular to $IP$ hence it must be $AQ$ . Since polars of $B,P,L$ concur, by dual we observe that $B,P,L$ are collinear, similarily $C,P,K$ are collinear. Thus, $Y=K$ and $Z=L$ . So $X$ is either $M_1$ or $M_2$ as desired. $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a