Nice tutorial problem for learning linearity! bgn wrote: Let $X,Y$ be two points on the side $BC$ of triangle $ABC$ such that $2XY=BC$ ($X$ is between $B,Y$). Let $AA'$ be the diameter of the circumcirle of triangle $AXY$. Let $P$ be the point where $AX$ meets the perpendicular from $B$ to $BC$, and $Q$ be the point where $AY$ meets the perpendicular from $C$ to $BC$. Prove that the tangent line from $A'$ to the circumcircle of $AXY$ passes through the circumcenter of triangle $APQ$. Proposed by Iman Maghsoudi Fix $\triangle AXY$ and move $B,C$ on line $\overline{XY}$ linearly, respecting $BC=2XY$. Then $P,Q$ also move linearly, along lines $\overline{AX}, \overline{AY}$; respectively. It suffices to show $\angle AA'O=90^{\circ}$ for precisely two cases. Claim. In triangle $AXY$; point $W$ is the reflection of $X$ in $Y$. Perpendicular at $W$ to $\overline{XY}$ meets $\overline{AY}$ at point $V$. Point $U$ is the circumcenter of $\triangle VAX$. Then $\angle AA'U=90^{\circ}$. (Proof) Let $M,N$ be mid-points of $\overline{XA}, \overline{VA}$; respectively. Drop perpendicular $\overline{AL}$ on $\overline{XY}$. Let $\angle XYA=y$ and $\angle AXY=x$. Note that $\tfrac{YV}{YA}=\tfrac{XY}{LY}$ so \begin{align*}\frac{XM}{YN}&=\frac{AX}{YV-YA} \\ &=\left(\frac{AX}{AY}\right) \cdot \left(\frac{LY}{XL}\right) \\ &= \left(\frac{\cos y}{\cos x}\right)\\ &=\left(\frac{A'X}{A'Y}\right) \end{align*}hence $\triangle A'XM \sim \triangle A'YN$ implying $A'$ lies on $\odot(AU)$, as claimed. $\blacksquare$ Note that the $B=X$ is equivalent to the claim; while $C=Y$ follows an analogous proof; hence the problem is solved. $\blacksquare$

Let $M,N$ be the midpoints of $AP,AQ$ respectively. Let $O$ be the circumcentre of $\triangle APQ$. We claim $\triangle A'MX \simeq \triangle A'NY$ We use the fact that $\triangle BPX$ is similar to $\triangle AA'Y$ and $\triangle CQY$ is similar to $\triangle AXA'$ and that $\angle AXM=\angle ANY=90^{\circ}$ which are obvious by angle chasing. $\triangle A'MX \simeq \triangle A'NY \Leftrightarrow \dfrac{MX}{A'X}=\dfrac{NY}{A'Y} \Leftrightarrow \dfrac{PX-XA}{A'X}=\dfrac{AY-QY}{A'Y} \Leftrightarrow \dfrac{PX}{A'X}+\dfrac{QY}{A'Y}=\dfrac{XA}{A'X}+\dfrac{AY}{A'Y} $ We now focus on the LHS: $LHS=\dfrac{BX \cdot \frac{AA'}{A'Y}}{A'X}+\dfrac{CY \cdot \frac{AA'}{A'X}}{A'Y}=\dfrac{AA' \cdot XY}{A'X \cdot A'Y}$ So $\triangle A'MX \simeq \triangle A'NY \Leftrightarrow \dfrac{AA' \cdot XY}{A'X \cdot A'Y}=\dfrac{XA}{A'X}+\dfrac{AY}{A'Y} \Leftrightarrow AA' \cdot XY=A'Y \cdot XA+AY \cdot A'X$ but this is just Ptolomey applied to cyclic quad $AXA'Y$ so the result is proved. Now we have $\measuredangle A'MA=\measuredangle A'NY=-\measuredangle A'NA=\measuredangle ANA'$ hence $AMA'N$ is cyclic. But also $AMON$ is obviously cyclic as $\angle OMA=\angle ONA=90^{\circ}$ hence $AMA'NO$ is cyclic with $AO$ as diameter so $\angle OA'A=90^{\circ}$ Hence $OA' \perp AA'$ but as $AA'$ is a diameter of circle $AXY$ this means $O$ is on the tangent to the circle at $A'$ as desired.

In my opinion this is easiest by trig. Notice that the problem is equivalent to proving $APQ$ and the reflection of $A$ over $A'$ are cyclic. (let that reflection be $S$. Here goes the trig: This is equivalent to $AP \cdot sin \angle SAQ +AQ \cdot sin \angle SAP = AS \cdot sin \angle PAQ$ $(*)$. Let $\angle PAQ=x$ (ptolomey) Note that $AS=4R$ where $R$ is the circumradius of $AXY$. So we need to prove $AS \cdot sin \angle SAQ + AQ \cdot sin \angle SAP = 4R \cdot sin x$ but obviously by extended Law of sines we have $\frac{XY}{sinx}=2R=\frac{BC}{2sinx}$ so we have $AS=4R=\frac{a}{sinx}$. So we need to prove $AP \cdot sin \angle SAQ +AQ \cdot sin \angle SAP = a$ $(**)$ Now let $AH$ be the altitude from $A$ to $BC$ (I do this purely for ease, not really needed). We have $\angle PAH= \angle SAQ=\angle APB$. Now we do law of sines on triangle $ABP$: $\frac{sin(90+\beta)}{AP}=\frac{sin(90-y)}{c}=\frac{cosy}{c}$ (here $y=\angle YXA$). Now after some fidgeting we have to prove $cos \beta \cdot c + cos \gamma \cdot b = a$ which is easy by Law of cosines.

Case $F$ $=$ $BC \cap PQ$ Let $O$, $H$ be the circumcenters of triangles $AXY$, $APQ$ respectively. We claim $H$$A^\prime$ is tangent to the circumcircle of $AXY$. We have $\angle HAA^\prime$ $=$ $\angle HAQ$ $-$ $\angle A^\prime AY$ $= 90^{\circ}$ $-$ $\angle APQ$ $-$ $\angle A^\prime AY$ but $\angle A^\prime AY$ $= 90^{\circ}$ $-$ $\angle Y A^\prime A$ $= 90^{\circ}$ $-$ $YXA$ $= 90^{\circ}$ $-$ $\angle BXP$ $=$ $\angle BPX$, so $\angle HAA^\prime$ $=$ $90^{\circ}$ $-$ $\angle APQ$ $-$ $\angle BPX$. Clearly $CQ \parallel BP$, so $\angle FCQ$ $=$ $\angle FBP$ and $\angle CFQ$ $= 90^{\circ}$$-$ $\angle APQ$ $-$ $\angle BPX$ $=$ $\angle HA A^\prime$ and $\frac{CF}{FQ}$ $=$ $\frac{BC}{PQ}$. On the other hand, $\angle XOY$ $=$ $\angle PHQ$ $=$ $2 \angle XAY$, so $\triangle XOY \sim \triangle PHQ$. From here, $\frac{XO}{XY}$ $=$ $\frac{AA^\prime}{BC}$ $=$ $\frac{PH}{PQ}$, from where, $\frac{AA^\prime}{PH}$ $=$ $\frac{BC}{PQ}$ $=$ $\frac{CF}{FQ}$, therefore $\frac{AA^\prime}{AH}$ = $\frac{CF}{FQ}$. Thus, $\triangle CFQ \sim \triangle A^\prime AH $, it follows that $\angle AA^\prime H$ $=$ $\angle OA^\prime H$ $= 90^{\circ}$, as desired $\blacksquare$ Case $BC \parallel PQ$ It's easy to see that $BCQP$ is a rectangle, from where $ 2 XY$ $=$ $PQ$, so $X$, $Y$ are the midpoints of $AP$, $AQ$, respectively. As $AA^\prime$ is diameter, it follows that $A^\prime$ is the intersection of the perpendicular bisectors of $AP$ and $AQ$, namely, $A^\prime$ is circumcenter of $\triangle APQ$ $\blacksquare$

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Let M and N be midpoints of AP and AQ. Let S,H,L,K be foots of M,A',A,N on BC. Let O be center of APQ. we need to prove AMA'NO is cyclic but it's easy to notice we can prove A'MX and A'YN are similar instead. we know A'HX and MSX are similar. AH'Y and YNK are similar as well. A'X/MX = A'H/XS and A'Y/NY = A'H/KY so we need to prove KY = XS. XS = BS - BX = BL/2 - BX KY = CY - CK = (BC/2 - BX) - CL/2 = BL/2 - BX we're Done.