# 2000 AMC 8 Problems/Problem 13

## Problem

In triangle $CAT$, we have $\angle ACT =\angle ATC$ and $\angle CAT = 36^\circ$. If $\overline{TR}$ bisects $\angle ATC$, then $\angle CRT =$ $[asy] pair A,C,T,R; C = (0,0); T = (2,0); A = (1,sqrt(5+sqrt(20))); R = (3/2 - sqrt(5)/2,1.175570); draw(C--A--T--cycle); draw(T--R); label("A",A,N); label("T",T,SE); label("C",C,SW); label("R",R,NW);[/asy]$ $\text{(A)}\ 36^\circ\qquad\text{(B)}\ 54^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 90^\circ\qquad\text{(E)}\ 108^\circ$

## Solution

In $\triangle ACT$, the three angles sum to $180^\circ$, and $\angle C = \angle T$ $\angle CAT + \angle ATC + \angle ACT = 180$ $36 + \angle ATC + \angle ATC = 180$ $2 \angle ATC = 144$ $\angle ATC = 72$

Since $\angle ATC$ is bisected by $\overline{TR}$, $\angle RTC = \frac{72}{2} = 36$

Now focusing on the smaller $\triangle RTC$, the sum of the angles in that triangle is $180^\circ$, so: $\angle RTC + \angle TCR + \angle CRT = 180$ $36 + \angle ACT + \angle CRT = 180$ $36 + \angle ATC + \angle CRT = 180$ $36 + 72 + \angle CRT = 180$ $\angle CRT = 72^\circ$, giving the answer $\boxed{C}$

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