Difference between revisions of "2023 USAMO Problems/Problem 1"

Revision as of 15:07, 16 April 2023

In an acute triangle $ABC$ , let $M$ be the midpoint of $\overline{BC}$ . Let $P$ be the foot of the perpendicular from $C$ to $AM$ . Suppose that the circumcircle of triangle $ABP$ intersects line $BC$ at two distinct points $B$ and $Q$ . Let $N$ be the midpoint of $\overline{AQ}$ . Prove that $NB=NC$ .

Solution 1

Let $X$ be the foot from $A$ to $\overline{BC}$ . By definition, $\angle AXM = \angle MPC = 90^{\circ}$ . Thus, $\triangle AXM \sim \triangle MPC$ , and $\triangle BMP \sim \triangle AMQ$ .

From this, we have $\frac{MP}{MX} = \frac{MA}{MC} = \frac{MP}{MQ} = \frac{MA}{MB}$ , as $MC=MB$ . Thus, $M$ is also the midpoint of $XQ$ .

Now, $NB = NC$ iff $N$ lies on the perpendicular bisector of $\overline{BC}$ . As $N$ lies on the perpendicular bisector of $\overline{XQ}$ , which is also the perpendicular bisector of $\overline{BC}$ (as $M$ is also the midpoint of $XQ$ ), we are done. ~ Martin2001, ApraTrip

@@ Line 8: / Line 8: @@
 ~ Martin2001, ApraTrip
 ==See also==
-{{USAMO box|year=2023|num-b=0|num-a=2|n=I}}
+{{USAMO box|year=2023|num-b=First Problem|num-a=2|n=I}}

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Difference between revisions of "2023 USAMO Problems/Problem 1"

Revision as of 15:07, 16 April 2023

Solution 1

See also