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Difference between revisions of "2000 AMC 8 Problems/Problem 13"

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Latest revision as of 23:36, 4 July 2013

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 (Problems • Answer Key • Resources)
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

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