1972 AHSME Problems/Problem 25

Inscribed in a circle is a quadrilateral having sides of lengths $25,~39,~52$ , and $60$ taken consecutively. The diameter of this circle has length

$\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad \textbf{(E) }69$

Solution

We note that $25^2+60^2=65^2$ and $39^2+52^2=65^2$ so our answer is $\boxed{C}$ .

-Pleaseletmewin

Alternate Solution: Let's call $\overline{AB}=25$ , $\overline{BC}=39$ , $\overline{CD}=52$ , $\overline{DA}=60$ . Let's call $\overline{BD}=x$ and $\angle{DAB}=y$ . By LoC we get the relation, $x^2=25^2+60^2-3000\cos(y)$ and $x^2=39^2+52^2-4056\cos(180-y)$ . If we do a bit of computation we get $x^2=4225-3000\cos(y)$ , and $x^2=4225-4056\cos(y)$ . This means that $4056\cos(y)=3000\cos(180-y)$ . We know that $\cos(180-y)=-\cos(y)$ so substituting back in we get $4056\cos(y)=-3000\cos(y)$ . We can clearly see that the only solution of this is $\cos(y)=0$ or $y=90$ . This then means that $\angle{BAD}=90$ and $\angle{BCD}=90$ . If a triangle is a right triangle and is inscribed in a circle then the diameter is the hypotenuse. This means that the diameter is $\sqrt{4225}=65$ so our answer is $\boxed{C}$ .

-Bole

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1972 AHSME Problems/Problem 25

Solution