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#1 h Apr 3, 2025, 12:19 AM • 3 Y

Y by sixoneeight, OronSH, goaoat

Quadrilateral $ABCD$ has an incenter $I$ Suppose $AB > BC$ . Let $M$ be the midpoint of $AC$ . Suppose that $MI \perp BI$ . $DI$ meets $(BDM)$ again at point $T$ . Let points $P$ and $Q$ be such that $T$ is the midpoint of $MP$ and $I$ is the midpoint of $MQ$ . Point $S$ lies on the plane such that $AMSQ$ is a parallelogram, and suppose the angle bisectors of $MCQ$ and $MSQ$ concur on $IM$ .

The angle bisectors of $\angle PAQ$ and $\angle PCQ$ meet $PQ$ at $X$ and $Y$ . Prove that $PX = QY$ .

This post has been edited 1 time. Last edited by popop614, Apr 3, 2025, 12:42 AM
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#2 h Apr 3, 2025, 12:42 AM

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popop614 wrote:

$MI \perp BI$ . $DI$

mibidi

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402 posts

Ianis

#3 h Apr 3, 2025, 12:45 AM

Y by

sixoneeight wrote:

popop614 wrote:

$MI \perp BI$ . $DI$

mibidi

skibidi?

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#4 h Apr 3, 2025, 3:28 AM

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Ianis wrote:

sixoneeight wrote:

popop614 wrote:

$MI \perp BI$ . $DI$

mibidi

skibidi?

dop dop yes yes

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goaoat

#5 h Apr 4, 2025, 8:00 PM • 2 Y

Y by mpcnotnpc, sixoneeight

sogood

:love:

$\textbf{Dealing with S:}$
First, let's eliminate $S$ . Note that the angle bisector concurrence is the same as $AQ/AM=MS/SQ=MC/CQ=AM/CQ \implies AM^2=AQ \cdot CQ$ Furthermore, the $PX=QY$ relation is the same as $AQ/AP=CP/CQ$ by angle bisector theorem, which we can simplify to $AM^2=AP \cdot CP$ using the previous equation. Thus, it suffices to prove this last relation.

$\textbf{Dealing with Q:}$
The equality $AM^2=AM \cdot AC=AQ \cdot CQ$ implies $Q$ lies on the lemniscate of Bernoulli with foci $A$ and $C$ , $\mathcal{L}$ . Thus, the inverse of $Q$ with respect to $(AC)$ lies on the rectangular hyperbola with foci $A$ and $C$ , $\mathcal{H}$ .

Now, the pedal curve of $\mathcal{H}$ is the image of $\mathcal{L}$ after a $\times 1/2$ homothety at $M$ . This gives that the perpendicular bisector of $MQ$ , or $BI$ , is tangent to $\mathcal{H}$ . However, since we have $\angle ABI = \angle CBI$ , by the optical property, the unique point on $BI$ that satisfies this is its tangency with $\mathcal{H}$ . Consequently, $B$ lies on $\mathcal{H}$ .

Now, $B$ lying on $\mathcal{H}$ is equivalent to $|AB-BC|=\frac{2+\sqrt{2}}{4} AC$ , and this can be rearranged with the median length formula to $AB \cdot BC = BM^2$ .

Let $Q'$ be the harmonic conjugate of $B$ with respect to $(BAC)$ . Then, $BC/BQ'=BM/AB \implies AB \cdot BC = BM \cdot BQ'=BM^2$ Thus, $BQ'=BM$ . Because $Q'$ also lies on the $B$ -symmedian, we conclude $Q=Q'$ so $Q$ lies on $(BAC)$ .

$\textbf{Dealing with D:}$
Furthermore, $ABCD$ is tangential so $AB+CD=AD+BC$ or $AB-BC=AD-CD$ , and this implies $D$ lies on $\mathcal{H}$ .

Claim: $\angle MBD = 90^\circ$

Proof: Let the reflection of $B$ over $M$ be $B'$ . Note $B'$ lies on $\mathcal{H}$ as $M$ is its center. Let $N$ be the midpoint of $BD$ . Realize $MN$ is the Newton line of $ABCD$ so $I$ lies on it. So, $BI \perp (B'D \parallel MN)$ . Since triangle $B'BD$ lies on $\mathcal{H}$ , its altitude $BI$ should intersect $\mathcal{H}$ at its orthocenter, but $BI$ is tangent to $\mathcal{H}$ . Ergo, $B$ is the orthocenter of triangle $B'BD$ , proving the claim.

$\textbf{Dealing with P:}$
The previous claim gives $\angle MTD = 90^\circ$ , so $P$ is the point on the $D$ -symmedian of $DAC$ satisfying $DM=DP$ .

Using the same phantom point argument we did with $Q$ , we can show that $P$ is the harmonic conjugate of $D$ with respect to $(DAC)$ .

Claim: The inverse of $P$ with respect to $(AC)$ is $D'$ , the reflection of $D$ over $AC$ .

Proof: It's easy to show that the inverse of $(DAC)$ with respect to $(AC)$ is $(D'AC)$ , the reflection of $(DAC)$ over $M$ . Let $D''$ be the reflection of $D$ over the perpendicular bisector of $AC$ . Note $D''$ is the intersection of $PD'$ with $(DAC)$ by some projective, so the inverse of $P$ is the reflection of $D''$ over $M$ which is $D'$ (it is not the reflection of $P$ over $M$ as they lie on different sides of $M$ ).

$\textbf{Finish:}$
Finally, $D'$ lies on $\mathcal{H}$ as $\mathcal{H}$ is symmetric about $AC$ , so $P$ lies on $\mathcal{L}$ . This means $AP \cdot CP = AM \cdot CM = AM^2$ , which is what we needed to prove, so we are done.

This post has been edited 3 times. Last edited by goaoat, Apr 4, 2025, 8:14 PM
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