Given an isosceles triangle $ ABC$ with $ \angle A = 90^{\circ}$. Let $ M$ be the midpoint of $ AB$. The line passing through $ A$ and perpendicular to $ CM$ intersects the side $ BC$ at $ P$. Prove that $ \angle AMC = \angle BMP$.
Problem
Source: Baltic Way 2000 Problem 2
Tags: trigonometry, geometry unsolved, geometry
dgreenb801
13.02.2009 04:40
Nice simple problem. Draw the parallel at B to AC, let it intersect AP at R. Then $ \triangle AMC \cong \triangle ABR$. Since $ \angle ABC=45$, it follows that $ \triangle BPM \cong \triangle BPR$, so $ \angle AMC = \angle BRA =\angle BMP$.
wangsacl
13.02.2009 05:00
dgreenb801 wrote: Nice simple problem. Draw the parallel at B to AC, let it intersect AP at R. Then $ \triangle AMC \cong \triangle ABR$. Since $ \angle ABC = 45$, it follows that $ \triangle BPM \cong \triangle BPR$, so $ \angle AMC = \angle BRA = \angle BMP$. Sorry, but I don't get it. Why $ \triangle AMC \cong\triangle ABR$? I don't get this fact... Edit: Stupid me, I don't recognize that $ ABC$ is a right isosceles triangle.
Sashsiam_2
13.02.2009 20:24
Note that $ \tan{M\hat{C}A}=\frac{1}{2}$, and that $ B\hat{A}P=M\hat{C}A$. Draw $ PQ$ perpendicular to $ AB$. Then we have$ PQ=x$$ =\frac{1}{2}AQ$, and as $ BPQ$ is similar to $ ABC$ then $ BQ=PQ=x$. So $ QM=AQ-AM=2x-1.5x=0.5x=\frac{1}{2}PQ$ then $ M\hat{P}Q=B\hat{A}P$ and the result follows.
sunken rock
04.11.2009 15:44
Take ABSC square, O it’s center. It’s obvious that $ \angle ABR =\angle CAM$, hence R is the middle of BS and P is the baricenter of $ \triangle ABS$, consequently M, P and S are collinear, but CM and SM are symmetrical w.r.t. OM, q.e.d. Best regards, sunken rock