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Winter Mathematical Competition

Geometry Difficulty 4.7 AIME Prove it Bulgaria

In acute $\triangle ABC$ denote by $M$ and $N$ the midpoints of the altitudes $BB_1$ and $CC_1$ , respectively, $P = AM \cap CC_1$ and $Q = AN \cap BB_1$ . Prove that:

a) the points $M, N, P$ and $Q$ are concyclic;

b) if the points $B, C, P$ and $Q$ are concyclic then $\triangle ABC$ is isosceles.

Solution

a) Since $\triangle ACC_1 \sim \triangle ABB_1$ and $AN$ and $AM$ are medians in these triangles we have
$\asymp ANC_1 = \asymp AMB_1 \Rightarrow \asymp QNB = \asymp PMQ,$
i.e. the points $M, N, P$ and $Q$ are concyclic.

b) If the points $B, C, P$ and $Q$ are concyclic then $\asymp QCP = \asymp QBP$ . But $\asymp ACC_1 = \asymp ABB_1$ and hence
$\asymp QCA = \asymp PBA. \qquad (1)$
On the other hand, since $\triangle ACC_1$ and $\triangle ABB_1$ are similar we have
$\asymp CAQ = \asymp CAN = \asymp BAM = \asymp BAP. \qquad (2)$
Now (1) and (2) imply that $\triangle ACQ \cong \triangle ABP$ , whence
$\frac{AC}{AB} = \frac{AQ}{AP} = \frac{AM}{AN} = \frac{AB}{AC}.$
Thus $AB^2 = AC^2$ and $AB = AC$ .

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