# 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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## Revision as of 11:20, 26 August 2020

Inscribed in a circle is a quadrilateral having sides of lengths , and taken consecutively. The diameter of this circle has length

## Solution

We note that and so our answer is .

-Pleaseletmewin

Alternate Solution: Let's call , , , . Let's call and . By LoC we get the relation, and . If we do a bit of computation we get , and . This means that . We know that so substituting back in we get . We can clearly see that the only solution of this is or . This then means that and . 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

-Bole