1972 AHSME Problems/Problem 25

Revision as of 11:20, 26 August 2020 by Bowenying24 (talk | contribs) (Solution)

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

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


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


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 $\sqrt{4225}=65$


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