silk road 2018 geometry

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Source: SRMC 2018 P1

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30629 posts

#1 h Aug 30, 2018, 11:09 PM • 3 Y

Y by mathematicsy, Adventure10, Mango247

In an acute-angled triangle $ABC$ on the sides $AB$ , $BC$ , $AC$ the points $H$ , $L$ , $K$ so that $CH \perp AB$ , $HL \parallel AC$ , $HK \parallel BC$ . Let $P$ and $Q$ feet of altitudes of a triangle $HBL$ , drawn from the vertices $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$ , drawn from the vertices $A$ and $H$ lie on the line $PQ$ .

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#2 h Aug 31, 2018, 9:08 AM • 2 Y

Y by Adventure10, Mango247

Let $D, E$ are feet of the altitudes of the triangle $AKH$ , drawn from the vertices $A$ and $H$
$\triangle AKH \sim \triangle ABC \sim \triangle HBL$ and $\triangle DEK \sim \triangle AHK$ , $\triangle QPL \sim \triangle BHL$ . So $DE \parallel PQ$
Easy to show, that $AE= AC \cos^2 \angle BAC,EC=AC \sin^2 \angle BAC$ and $PC=BC \sin^2 \angle ABC$
$\frac{EC}{PC}=\frac{AC \sin^2 \angle BAC}{BC \sin^2 \angle ABC}= \frac{BC}{AC}=\frac{BL}{HL}=\frac{LQ}{LP}=\frac{EK}{KD}$ so $P,Q,D,E$ lies on same line.

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#3 h Apr 7, 2019, 5:37 PM • 3 Y

Y by NikitosKh, Adventure10, Mango247

Let $M, N$ are feet of the altitudes of the triangle $AKH$ , drawn from the vertices $A$ and $H$ respectively.
Since $HC$ is a radical axis of the circles, which are circumscribed around $ANMH$ and $HQPB$ , and $C$ belongs $HC$ we get:
$CP*CB=CN*CA$ So, the quadrilateral $ANPB$ is cyclic and $\angle BAC = \angle NPC$ . We know that
$\triangle AKH \sim \triangle ACB \sim \triangle HLB$ and we get $\angle BAC = \angle NMK=\angle QPL=\angle QPC$ . So $\angle NPC = \angle QPC$ and we get that $P,Q,N$ lies on the same line.
Similarly we can prove that $P$ belongs $MN$ . And since $Q$ belongs $NP$ we are done.

This post has been edited 2 times. Last edited by Daniil02, Apr 7, 2019, 5:39 PM

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#4 h May 14, 2019, 6:59 PM • 2 Y

Y by Adventure10, Mango247

feet of altitude from miquel point are collinear

This post has been edited 1 time. Last edited by AlastorMoody, May 14, 2019, 7:00 PM

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#5 h Jul 20, 2019, 1:08 PM • 2 Y

Y by top1csp2020, Adventure10

Let AE, HD perp KE, KH and F is orthocentre hence HEQF cylic so easy angle chase so done

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#6 h Aug 18, 2023, 5:42 PM

Y by

Let the foots of perpendicular from $H$ and $A$ to $AK$ and $HK$ are $Z$ and $Y$ , respectively. Then $(H Z C P)$ is cyclic, then we have $\angle HPZ=\angle HCA=90- \angle A=90-\angle BHQ=\angle HBQ=\angle HPQ$ , which means that $Z$ lies on $PQ$ , similarly we can prove that $Y$ lies on $PQ$ (let $BQ$ and $AY$ intersect at $X$ , use that $(H Q X Y)$ cyclic).

This post has been edited 2 times. Last edited by X.Allaberdiyev, Aug 18, 2023, 5:43 PM

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