# Proving the line is tangent to the circle (solution APMO 2016)

Proving the line is tangent to the circle (solution APMO 2016)

Problem 3.

Let AB and AC be two distinct rays not lying on the same line, and let be a circle with Center O that is tangent to ray AC at E and ray AB at F. Let R be a point on segment EF. The line through O parallel to EF intersects line AB at P. Let N be the intersection of lines PR and AC, and let M be the intersection of line AB and the line through R parallel to AC. Prove that line MN is tangent to .

Also Read:

(**APMO 2016, no. 3)**

**Solution.** We present two approaches. The first one introduces an auxiliary point and studies similarities in the figure. The second one reduces the problem to computations involving a particular exradius of a triangle. The second approach has two variants.

**Solution 1.**

Let the line through N tangent to ! at point intersect AB at point M’. It suffices to show that M’R || AC, since this would yield M’ = M.

Suppose that the line PO intersects AC at Q and the circumcircle of AM’O at Y, Respectively. Then

.

By angle chasing we have and by symmetry . Therefore .

On the other hand, we have

Since we know that and are complementary this implies

Therefore, and are congruent angles, and this means that A and N are corresponding points in the similarity of triangles and . It follows that

We conclude that M’R || AC, as desired.

**Also Read : Solution Combinatoric Problem in APMO ( Solution APMO 2016 number 4)**

**Solution 2.**

As in Solution 1, we introduce point M’. and reduce the problem to proving . Menelaus theorem in triangle ANP with transversal line FRE yields

.

Since AF = EA, we have , so that it suffices to prove

This is a computation regarding the triangle AM’N and its excircle opposite A. Indeed, setting *a = *M’N, b = NA, c = M’A, , x = s – a, y = s – b, and z = s – c, then AE = AF = s, M’F = z, and NE = y. from we have , where r_{a} = OF is the exradius opposite A. Combining the following two standard formulas for the area of a triangle

|AM’N|^{2} = xyzs (Heron’s Formula) and |AM’N| = r_{a} (s – a).

We have , therefore, , We can now write everything in (1) in terms of x, y, z. We conclude that we have to verify

which is easily seen to be true.

**Note**: Another approach using Menalaus theorem is to construct the tangent from M to create a point N’ in AC and then prove, using the theorem, that P, R and N’ are collinear. This also reduces to an algebraic identity.