tangent circumcircles wanted, altitudes, orthocenter related

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Source: Ukraine Online MO 2020 Grade 10.3

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#1 h Sep 7, 2020, 7:28 PM • 2 Y

Y by Kimchiks926, Rounak_iitr

Let $H_a,H_b,H_c$ be feet of altitudes from the corresponding vertices of the triangle $ABC, H$ be the orthocenter of this triangle, and $K$ is a point that is symmetric to $H$ wrt $BC$ . A line passing through point $H$ parallel to $H_bH_c$ , intersects line $AB$ and $AC$ at points $X$ and $Y$ respectively. . Prove that the circles are circumscribed to $\vartriangle ABC$ and $\vartriangle XYK$ are tangent.

(Bondarenko Mykhailo)

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#2 h Sep 30, 2020, 5:42 PM

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Wlog $AB<AC$ . Note that $BCXY$ is cyclic since $XY||H_bH_c$ and $BCH_bH_c$ . Now let $D=XY \cap BC$ . By angle-chasing we have,
$\angle DKA=\angle DKH=\angle XHH_a=\angle XHB+\angle BHH_a=\angle H_cH_bB+\angle BH_cH_a$
$\angle ACK=\angle BCK+\angle C=\angle H_cH_bB+\angle BH_cH_a$
So $DK$ is tangent to $(ABC)$ and by converse Radixal axis theorem on $(ABCK),(BCXY),(KXY)$ we get that $DK$ is tangent to $(XYK)$ as well

This post has been edited 2 times. Last edited by Com10atorics, Dec 31, 2020, 4:32 PM

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#3 h Oct 1, 2020, 1:14 AM

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Let $XY$ intersects $BC$ at $Z$ then $\angle{ZHB} = \angle{CAO} = \angle{ACO} = \angle{HCB}$ (With $O$ is center of $(ABC)$ ). So $ZK^2 = ZH^2 = \overline{ZB} \cdot \overline{ZC} = \overline{ZX} \cdot \overline{ZY},$ which means $ZK$ is common tangent of $(ABC), (XYK)$

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#4 h Dec 15, 2020, 6:21 PM • 2 Y

Y by Mango247, Mango247

Let $L$ and $M$ be the reflections of $H$ w.r.t. $\overline{CA}$ , $\overline{AB}$ , respectively. It's well-known that $K$ , $L$ , $M$ are on the circumcircle of $\triangle ABC$ . Let $\overline{KL}$ meet $\overline{CA}$ at $Z$ . Then
$\angle AZH=180^{\circ}-\angle HAZ - \angle AHZ=180^{\circ}-\angle HAH_B-\angle AHH_B-\angle H_BHZ=90^{\circ}-\angle H_BHZ=90^{\circ}-\angle H_BLZ=90^{\circ}-\angle BCK=90^{\circ}-\angle BAK=90^{\circ}-(90^{\circ}-\angle ABC)=\angle ABC= \angle AH_BH_C \implies \overline{HZ} \parallel \overline{H_BH_C} \implies Z \equiv Y \implies K,Y,L$ are collinear. Similarly $K,X,M$ are collinear. Notice that from the midlines $\overline{ML} \parallel \overline{H_BH_C} \parallel \overline{XY} \implies \triangle MKL$ and $\triangle XKY$ are homothetic with center $K$ , thus their circumcircles are tangent, as desired.

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