# 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 $\omega$ 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 $\omega$.

(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 $X\neq E$ 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

$\angle AYM'=\angle AOM'=90^0-\angle M'OP$.

By angle chasing we have $\angle EOQ=\angle FOP=90^0-\angle AOF=\angle M'AO=\angle M'YP$ and by symmetry $\angle EQO=\angle M'PY$ . Therefore $\Delta M'YP\sim \Delta EOQ$.

On the other hand, we have

Since we know that $\angle AYM'$ and $\angle M'OP$ are complementary this implies

$\angle AYM'=\frac{\angle XOE}{2}=\angle NOE$

Therefore, $\angle AYM'$ and $\angle NOE$ are congruent angles, and this means that A and N are corresponding points in the similarity of triangles $\Delta M'YP$ and $\Delta EOQ$. It follows that

$\frac{AM'}{M'P}=\frac{NE}{EQ}=\frac{NR}{RP}$

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 $\frac{PR}{RN}=\frac{PM'}{M'A}$. Menelaus theorem in triangle ANP with transversal line FRE yields

$\frac{PR}{RN}.\frac{NE}{NA}.\frac{AF}{FP}=1$.

Since AF = EA, we have $\frac{FP}{NE}=\frac{PR}{RN}$, so that it suffices to prove

$\frac{FP}{NE}=\frac{PM'}{M'A}$

This is a computation regarding the triangle AM’N and its excircle opposite A. Indeed, setting a = M’N, b = NA, c = M’A, $s=\frac{a+b+c}{2}$, x = s – a, y = s – b, and z = s – c, then AE = AF = s, M’F = z, and NE = y. from $\Delta OFP\sim \Delta AFO$ we have $FP=\frac{r_a^2}{s}$, where ra = 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| = ra (s – a).

We have $r_a^2=\frac{yzs}{x}$, therefore, $FP=\frac{yz}{x}$, 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.