## Monday, February 6, 2006

### Meeting Time! Topic: Geometry. Level: Olympiad.

Problem: (1997 Hungary-Israel, MOP Geometry Packet) The three squares $ACC_1A^{\prime\prime}$, $ABB_1A^{\prime}$, $BCDE$ are constructed externally on the sides of triangle $ABC$. Let $P$ be the center of $BCDE$. Prove that the lines $A^{\prime}C, A^{\prime\prime}B, PA$ are concurrent.

Solution: So a fun concurrency problem for the day... first off, we must remember that there are a bunch of right angles in there because of squares and I'm not going to be explicitly saying it each time, but whenever I assume an angle to be right, it will be in a square. Let $X$ be the intersection of $BA^{\prime\prime}$ and $CA^{\prime}$ (not on diagram). Note that we DO NOT KNOW if $X$ lies on $AP$ yet (or we would have assumed the result). $F$ is the extension of $AX$ to meet $BC$. We do not know if $P$ lies on line $AXF$.

First, look at $\triangle BAA^{\prime\prime}$ and $\triangle CAA^{\prime}$. We see that $CA = AA^{\prime\prime}$, $BA = AA^{\prime}$, and $\angle BAA^{\prime\prime} = 90^{\circ}+\angle BAC = \angle CAA^{\prime}$ so by SAS congruency, we have the two triangles congruent. Hence $\angle AA^{\prime}H = \angle XBH$ and $\angle AA^{\prime\prime}G = XCG$.

Then $\angle A^{\prime}XB = 180^{\circ}-(45^{\circ}-\angle AA^{\prime}H)-(45^{\circ}+\angle XBH) = 90^{\circ}$. But since $\angle A^{\prime}AB = 90^{\circ}$ as well, quadrilateral $A^{\prime}AXB$ is cyclic.

From that, we have $\angle BAX = \angle BA^{\prime}X = 45^{\circ}-\angle AA^{\prime}H = 45^{\circ}-\angle XBH = 45^{\circ}-\angle XBA$, so $\angle AXB = 180^{\circ}-\angle BAX-\angle XBA = 135^{\circ}$. That gives us $\angle BXF = 45^{\circ}$. Similarly, for the other side, we can show that $\angle CXF = 45^{\circ}$ and $\angle BXC = 90^{\circ}$.

Then consider quadrilateral $XBCP$. Since $\angle BXC+\angle BPC = 90^{\circ}+90^{\circ} = 180^{\circ}$, we have $XBCP$ cyclic. Therefore, $\angle BXP = \angle BCP = 45^{\circ}$. But since $\angle BXF = 45^{\circ}$ as we found above, $P$ must lie on line $AXF$. Therefore, $A, X, F, P$ are collinear and $X$, the intersection of $BA^{\prime\prime}$ and $CA^{\prime}$, lies on $AP$, meaning $BA^{\prime\prime}, CA^{\prime}, AP$ are concurrent, as desired. QED.

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Comment: Concurrency problems (showing that three or more lines intersect at a single point) can be solved in a variety of ways. One of them, shown in this problem, is proving that the intersection of two lines lies on the third. Other methods include Ceva's Thoerem and the trigonometric from of Ceva's Theorem. Another topic that goes along with this nicely is collinearity - showing three or more points lie on a single line.

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Practice Problem: (1997 USAMO - #2) Let $ABC$ be a triangle, and draw isosceles triangles $DBC,AEC,ABF$ external to $ABC$ (with $BC,CA,AB$ as their respective bases). Prove that the lines through $A,B,C$ perpendicular to $EF, FD,DE$, respectively, are concurrent.