1959 AHSME Problems/Problem 36
Problem
The base of a triangle is 
, and one side of the base angle is 
. The sum of the lengths of the other two sides is 
. The shortest side is:
Solution
![[asy] import geometry; size(10cm); point A = (0,0); point B = (17/20,17*sqrt(3)/20); point C = (8,0); triangle ABC = triangle(A,B,C); // Triangle ABC draw(ABC); dot(A); label("A",A,SW); dot(B); label("B",B,NW); dot(C); label("C",C,SE); // Labels markscalefactor = 0.1; draw(anglemark(C,A,B)); label("$60^{\circ}$", A, (1.5,1.5)); label("$80$", midpoint(A--C), S); label("$x$", midpoint(A--B), NW); label("$90-x$", midpoint(B--C), NE); [/asy]](http://latex.artofproblemsolving.com/4/1/5/41565499f5c6d0c9ea191667c4def91b495fadad.png)
Let the triangle be 
with 
and 
, as in the diagram. Let 
. from the problem, we know that 
. Now, we can apply the Law of Cosines on to solve for 
:
\begin{align*}
(90-x)^2 &= 6400+x^2-160\cos(60^{\circ}) \\
x^2-180x+8100 &= 6400+x^2-80x \\
-100x &= -1700 \\
x &= 17
\end{align*}
Because the problem asks for the shortest side of the triangle and 
, our answer is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 35 |
Followed by Problem 37 | |
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
