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 $\boxed{C}$

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png