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Source: PErA 2025/4

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#1 h Mar 17, 2025, 1:50 PM

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Let $ABC$ be an acute-angled scalene triangle. Let $B_1$ and $B_2$ be points on the rays $BC$ and $BA$ , respectively, such that $BB_1 = BB_2 = AC$ . Similarly, let $C_1$ and $C_2$ be points on the rays $CB$ and $CA$ , respectively, such that $CC_1 = CC_2 = AB$ . Prove that if $B_1B_2$ and $C_1C_2$ intersect at $K$ , then $AK$ is parallel to $BC$ .

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#2 h Mar 17, 2025, 6:06 PM • 2 Y

Y by HotSinglesInYourArea, Calamarul

Really short bary solution.

We have $B_1=(0:a-b:b)$
$B_2=(b:c-b:0)$
$C_1=(0:c:a-c)$
$C_2=(c:0:b-c)$

Take $\ell$ to be the parallel to $BC$ through $A$ and consider $X=B_1B_2\cap \ell$ and $Y=C_1C_2\cap \ell$

Then $X=(1,x,-x)$ and $Y=(1,y,-y)$ for some reals $x, y$ .

Since $B_1, B_2, X$ are collinear we have $\det \begin{bmatrix} b&c-b&0\\ 0&a-b&b\\ 1&x&-x \end{bmatrix}=0\iff -bx(a-b)+b(c-b)-b^2x=0\iff ax=c-b\iff x=\frac{c-b}{a}$ .

Compute $y=\frac{c-b}{a}$ similarly and conclude that $X=Y$ .

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