Let $ABC$ be an isosceles triangle, $AB=AC$ inscribed in a circle $\omega$. The $B$-symmedian intersects $\omega$ again at $D$. The circle through $C,D$ and tangent to $BC$ and the circle through $A,D$ and tangent to $CD$ intersect at points $D,X$. The incenter of $ABC$ is denoted $I$. Prove that $B,C,I,X$ are concyclic.
Problem
Source: 2023 Israel TST Test 5 P2
Tags: geometry, TST, symmedian, incenter
trinhquockhanh
24.03.2023 06:01
My solution with angle chasing and similar triangles, the critical step is to construct $T$ and $Y$.
Hermione.Potter
26.03.2023 05:18
Let $(CDX) \cap AC = M$. Claim: $M$ is the midpoint of $AC$. Proof: $\angle CDM = \angle BCA = \angle BDA$ and $BD$ is the $D$-symmedian of $ACD$, thus $DM$ must be the median. Claim: $BXMA$ is cyclic. Proof: This follows by angle chase: $$\angle MBA = \angle CBD = \angle CAD = \angle XAD - \angle XAM = \angle XDC - \angle XAM = \angle XMC - \angle XAM = \angle MXA$$ To finish, note that: $$\angle BXC = 360^{\circ} - \angle CXM - \angle BXM = 360^{\circ} - (180^{\circ} - \angle CDM) - (180^{\circ} - \angle BAC) = \angle CDM + \angle BAC = \angle CBA + 180^{\circ} - 2\angle CBA = 180^{\circ} - \angle CBA = \angle BIC,$$which completes the proof. Comment: Additionally, $C, Y, X $ are collinear, and $DY, BM$ should meet on $(CDX)$. ($Y = AB \cap (ADX)$)
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ZNatox
27.03.2023 01:26
Symmedians, the worst of them all. Anyway, Let $M$ be the midpoint of $AC$. It is well-known that $X \in (MDC)$. Now, $\angle CXB= 2\pi - \angle CXD - \angle DXA - \angle AXB = (\pi-\angle CXD) + (\pi - \angle DXA) - \angle AXB = (\angle AMB) + (\pi - \angle CBA) - \angle AXB \\ = \frac{\pi}{2} + \frac{\angle A}{2} + \angle AMB - \angle AXB = \angle BIC + \angle AMB - \angle AXB$. So, we must show $X \in (AMB)$. For this, we use $\sqrt{ac}$ inversion centered at $B$. Let $(ADX)\longleftrightarrow \omega, X\longleftrightarrow X_1$. Notice that $(AMB)$ and $\omega$ are tangent and that $C,M \in \omega$. We have $\angle ADX_1= \angle AMX_1= \angle ABM + \angle ACD = \angle B$, which implies $X_1 \in (CD)$ and we're done .
crazyeyemoody907
28.03.2023 01:34
By design, $ABCD$ is harmonic, so there exists a (unique) Brokard point. To see that $X$ is this point, note that it satisfies the condition that three of the angles $XAD, XDC, XCB$ are equal, and only the Brokard point satisfies these. Having concluded that $X$ is the Brokard point of harmonic quadrilateral $ABCD$, we have $\angle XBA=\angle XCB$, so $\overline{AB}$ is tangent to $(XBC)$; it's well-known that the only circle tangent to $\overline{AB}$ at $B$ and passing through $C$ is $(BIC)$ for isosceles triangle $ABC$, so we are done! A...D....C....D