Belorusija, 2012

Geometry Difficulty 6.8 National Olympiad Prove it Belarus

Two squares with the centers $C_1$ and $B_1$ are constructed on the sides $AB$ and $AC$ outside of the acute-angled triangle $ABC$ , respectively. The square $C_1B_1DE$ is constructed on the segment $C_1B_1$ so that $A$ and $D$ lie in the different half-planes with respect to $C_1B_1$ .
Prove that the center of the square $C_1B_1DE$ belongs to the line $BC$ .

Solution

Let $H$ be a foot of perpendicular from $A$ onto $BC$ . Since $\angle AB_1C = 90^\circ$ , we have $\angle AHC + \angle AB_1C = 90^\circ + 90^\circ = 180^\circ$ . Thus, points $A, B_1, C, H$ lie on the same circle. Since $AB_1 = B_1C$ , we see that $HB_1$ is a bisector of $\angle AHC$ , so $\angle AHB_1 = \angle B_1HC = 90^\circ/2 = 45^\circ$ . In the same way we obtain $\angle AHC_1 = \angle C_1HB = 45^\circ$ .

Further, consider the circle $\omega$ with the diameter $C_1B_1$ . Since $\angle C_1HB_1 = \angle C_1HA + \angle AHB_1 = 45^\circ + 45^\circ = 90^\circ$ , we obtain $H \in \omega$ . Let $T$ be the second point of intersection of $\omega$ and $BC$ . Then $\angle C_1B_1T = \angle C_1HT = 45^\circ$ and $\angle TC_1B_1 = \angle B_1HC = 45^\circ$ . Therefore, $\triangle C_1TB_1$ is an isosceles right-angled triangle, so point $T \in BC$ coincides with the center of the square $C_1B_1DE$ .

In the case when point $T$ coincides with $H$ , i.e. $\omega$ touches $BC$ , we have $\angle C_1B_1H = \angle C_1HB = 45^\circ$ and $\angle B_1C_1H = \angle B_1HC = 45^\circ$ . Therefore, point $H$ is the center of the square $C_1B_1DE$ .

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