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Source: Saudi Arabia JBMO Training test 1 2019, P2

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#1 h Apr 19, 2020, 4:31 PM

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Let $AA_1$ and $BB_1$ be heights in acute triangle intersects at $H$ . Let $A_1A_2$ and $B_1B_2$ be heights in triangles $HBA_1$ and $HB_1A$ , respe. Prove that $A_2B_2$ and $AB$ are parralel.

This post has been edited 1 time. Last edited by JBMO2020, Apr 19, 2020, 4:33 PM

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brainiacmaniac31

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brainiacmaniac31

#2 h Apr 19, 2020, 10:35 PM

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If we can show that $\triangle A_2HB_2\sim\triangle BHA$ , then we know that $A_2B_2\parallel AB$ .
$[asy] import olympiad; import geometry; size(250); currentpen=fontsize(9); pair a,b,c,h,a1,b1,a2,b2; a=dir(105); b=dir(-90-50); c=dir(-90+50); h=a+b+c; a1=foot(a,b,c); b1=foot(b,a,c); a2=foot(a1,b,b1); b2=foot(b1,a,a1); draw(a--b--c--a--a1--a2^^b--b1--b2--a2); label("$A$",a,dir(a)); label("$B$",b,dir(b)); label("$C$",c,dir(c)); label("$H$",h,dir(-30)); label("$A_1$",a1,dir(a1)); label("$A_2$",a2,dir(180-40)); label("$B_1$",b1,dir(b1)); label("$B_2$",b2,dir(b2)); [/asy]$
In triangle $BA_1H$ , $A_1A_2$ is an altitude, so $\triangle BA_1H\sim \triangle A_1A_2H$ . This implies that $\frac{A_2H}{A_1H}=\frac{A_1H}{BH}$ , or that $A_2H=\frac{A_1H^2}{BH}$ . Similarly, $B_2H=\frac{B_1H^2}{AH}$ . In order to show that $\triangle A_2HB_2\sim\triangle BHA$ , we will show that $\frac{A_2H}{B_2H}=\frac{BH}{AH}$ . By virtue of our previous calculations, we know that
$\frac{A_2H}{B_2H}=\frac{A_1H^2/BH}{B_1H^2/AH}=\left(\frac{A_1H}{B_1H}\right)^2\cdot\frac{AH}{BH}.$ On the other hand, since $AB_1A_1B$ is cyclic, by Power of a Point we know that $(AH)(A_1H)=(BH)(B_1H)\implies \frac{A_1H}{B_1H}=\frac{BH}{AH}$ . Thus,
$\frac{A_2H}{B_2H}=\left(\frac{A_1H}{B_1H}\right)^2\cdot\frac{AH}{BH}=\left(\frac{BH}{AH}\right)^2\cdot\frac{AH}{BH}=\frac{BH}{AH},$ which implies that $\triangle A_2HB_2\sim\triangle BHA$ as desired.

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#3 h Apr 19, 2020, 10:55 PM

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Since $AB$ and $A_1B_1$ are antiparallel in $\angle AHB,$ it suffices to show that $A_2B_2$ and $A_1B_1$ are also antiparallel in the same angle. This just follows from the fact that $\angle A_1A_2B_1=\angle A_1B_2B_1=90^\circ,$ so $A_1A_2B_2B_1$ is cyclic. $\blacksquare$

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#5 h Jul 30, 2020, 5:10 AM • 3 Y

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Note that $A_1B_1A_2B_2$ is cyclic since $\angle A_1B_2B_1= 90^\circ=\angle A_1A_2B_1$ . It is well known that so is $AA_1BB_1$ . So by angle chasing we have,
$\angle B_2A_2H=\angle B_2A_2B_1=\angle B_2A_1B_1=\angle AA_1B_1=\angle ABB_1=\angle ABH$

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jayme

#6 h Jul 30, 2020, 12:32 PM

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Dear Mathlinkers,

why not with the Reim's theorem?

Sincerely
Jean-Louis

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jayme

#7 h Sep 17, 2020, 12:16 PM

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Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%209.pdf p. 58-59.

Sincerely
Jean-Louis

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#8 h Feb 28, 2023, 8:14 AM

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solution with formal sum.

$A_2B_2+A_1B_1=A_1H+B_1H=AH+BH=AB+A_1B_1$
So $AB=A_2B_2$ . We're done.

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