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Difference between revisions of "1972 AHSME Problems/Problem 25"

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

Revision as of 12:21, 26 August 2020

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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