Moldova TST 2005 for JBMO, day 1, problem 1

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Moldova TST 2005 for JBMO, day 1, problem 1

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Source: Moldova TST 2005 for JBMO, day 1, problem 1

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orl

#1 h Apr 11, 2005, 5:34 PM • 2 Y

Y by Adventure10, Mango247

Let the triangle $ABC$ with $BC$ the smallest side. Let $P$ on ( $AB$ ) such that angle $PCB$ equals angle $BAC$ . and $Q$ on side ( $AC$ ) such that angle $QBC$ equals angle $BAC$ . Show that the line passing through the circumenters of triangles $ABC$ and $APQ$ is perpendicular on $BC$ .

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#2 h Apr 11, 2005, 6:19 PM • 2 Y

Y by Adventure10, Mango247

In general, let $X'$ be the symmetric of $X$ wrt the perpendicular bisector of $BC$ . If we show that $A'$ lies on the circle $(APQ)$ , then we're done, because $A,A'$ also lie on $(ABC)$ , so the line of the centers of $(APQ),(ABC)$ is $\perp AA'$ , which is $\|BC$ (I'm assuming $AB\ne AC$ , the other case is trivial).

We have $\angle PQ'Q=\pi-\angle QQ'C=\pi-\angle A$ , so $Q'$ lies on $(APQ)$ . In basicly the same way we show that $P'\in(APQ)$ . Because of the symmetry of the situation, we also know that $Q\in(A'P'Q')$ . We have thus shown that both $A,A'$ lie on $(PQQ')$ , and this proves that $APQA'$ is cyclic.

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#3 h Apr 12, 2005, 8:27 AM • 2 Y

Y by Adventure10, Mango247

orl wrote:

Let the triangle $ABC$ with $BC$ the smallest side. Let $P$ on ( $AB$ ) such that angle $PCB$ equals angle $BAC$ . and $Q$ on side ( $AC$ ) such that angle $QBC$ equals angle $BAC$ . Show that the line passing through the circumenters of triangles $ABC$ and $APQ$ is perpendicular on $BC$ .

Another solution:

In the following, the abbreviation "circle M(r)" will mean the circle with the center M and the radius r, and the abbreviation "circle $P_1P_2P_3$ " will mean the circle through three given points $P_1$ , $P_2$ , $P_3$ .

We have to show that the line joining the centers of the circles ABC and APQ is perpendicular to the line BC.

Since < QBC = < BAC and < QCB = < BCA, the triangles QBC and BAC are similar. Thus, CA : CB = CB : CQ. In other words, $CA\cdot CQ=CB^2$ . Since the point Q lies on the ray CA, we can therefore conclude that the points A and Q are inverse to each other with respect to the circle C(CB). Now, if a circle u passes through two points which are inverse to each other with respect to another circle k, then this circle u is orthogonal to the circle k. Hence, the circle APQ, passing through the points A and Q which are inverse to each other with respect to the circle C(CB), must be orthogonal to this circle C(CB). Similarly, the circle APQ is also orthogonal to the circle B(BC). But the centers of all circles orthogonal to two given circles lie on the radical axis of these two circles. Hence, since the circle APQ is orthogonal to the two circles B(BC) and C(CB), its center lies on the radical axis of these two circles B(BC) and C(CB). Now, clearly, these two circles B(BC) and C(CB) have equal radius, and since the radical axis of two circles with equal radius is the perpendicular bisector of the segment joining their centers, it follows that the radical axis of the circles B(BC) and C(CB) is the perpendicular bisector of the segment BC. Hence, we obtain that the center of the circle APQ lies on the perpendicular bisector of the segment BC. On the other hand, it is trivial that the center of the circle ABC also lies on the perpendicular bisector of the segment BC. Hence, the line joining the centers of the circles ABC and APQ is the perpendicular bisector of the segment BC, and thus it is perpendicular to the line BC. Problem solved.

Darij

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#4 h Aug 4, 2023, 2:02 PM

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Let $K$ be the intersection of $CP$ and $(APQ)$ (other than $P$ ). It's easy to see that $KQ \parallel BC$ , because $\angle QKC= \angle BAC= \angle BCK$ .
We have $\angle KCB = \angle QBC$ , so $BCQK$ is a isosceles trapezoid.
$I$ , $Q$ lies on the perpendicular bisector of $KQ$ , $BC$ respectively. But $KQ$ , $BC$ have the same perpendicular bisector, so we are done.

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