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

(Created page with "==Solution== We note that <math>25^2+60^2=65^2</math> and <math>39^2+52^2=65^2</math> so our answer is <math>\boxed{65}</math>. -Pleaseletmewin") |
Coolmath34 (talk | contribs) m (→Solution) |
||

(3 intermediate revisions by 2 users not shown) | |||

Line 1: | Line 1: | ||

+ | Inscribed in a circle is a quadrilateral having sides of lengths <math>25,~39,~52</math>, and <math>60</math> taken consecutively. The diameter of this circle has length | ||

+ | |||

+ | <math>\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad \textbf{(E) }69</math> | ||

+ | |||

==Solution== | ==Solution== | ||

− | We note that <math>25^2+60^2=65^2</math> and <math>39^2+52^2=65^2</math> so our answer is <math>\boxed{ | + | We note that <math>25^2+60^2=65^2</math> and <math>39^2+52^2=65^2</math> so our answer is <math>\boxed{C}</math>. |

-Pleaseletmewin | -Pleaseletmewin | ||

+ | |||

+ | Alternate Solution: | ||

+ | |||

+ | 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 relations | ||

+ | |||

+ | <cmath>x^2=25^2+60^2-3000\cos(y)</cmath> | ||

+ | <cmath>x^2=39^2+52^2-4056\cos(180-y)</cmath> | ||

+ | |||

+ | 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 (edited for easier readability) |

## Latest revision as of 09:14, 29 January 2021

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 relations

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 so our answer is .

-Bole (edited for easier readability)