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1998 AHSME Problems/Problem 26

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Problem

In quadrilateral $ABCD$ , it is given that $\angle A = 120^{\circ}$ , angles $B$ and $D$ are right angles, $AB = 13$ , and $AD = 46$ . Then $AC=$

$\mathrm{(A)}\ 60\qquad\mathrm{(B)}\ 62\qquad\mathrm{(C)}\ 64\qquad\mathrm{(D)}\ 65\qquad\mathrm{(E)}\ 72$

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

Solution

Solution 1

Let the extensions of $\overline{DA}$ and $\overline{CB}$ be at $E$ . Since $\angle BAD = 120^{\circ}$ , $\angle BAE = 60^{\circ}$ and $\triangle ABE$ is a $30-60-90$ triangle. Also, $\triangle ABE \sim \triangle CDE$ , so $\triangle CDE$ is also a $30-60-90$ triangle.

size(200);defaultpen(0.8);pair D=(0,0), C=(0,24*3^0.5), A=(46,0), E=(72,0), B=(46+13/2,13*3^.5/2);pair P=(C+D)/2, Q=(D+A)/2, ...

Thus $AE = 2AB = 26$ , and $CD = \frac{26 + 46}{\sqrt{3}} = 24\sqrt{3}$ . By the Pythagorean Theorem on $\triangle ACD$ , $AC = \sqrt{(46)^2 + (24\sqrt{3})^2} = 62 \Rightarrow \mathrm{(B)}.$

Solution 2

import olympiad;size(180);defaultpen(0.8);pair D=(0,0), C=(0,24*3^0.5), A=(46,0), B=(46+13/2,13*3^.5/2);pair P=(C+D)/2, Q=(D+...

Opposite angles add up to $180^{\circ}$ , so $ABCD$ is a cyclic quadrilateral. Also, $\angle B = \angle D = 90^{\circ}$ , from which it follows that $\overline{AC}$ is a diameter of the circumscribing circle. We can apply the extended version of the Law of Sines on $\triangle ABD$ :

$AC = 2R = \frac{BD}{\sin 120^{\circ}} = \frac{2}{\sqrt{3}} BD$

By the Law of Cosines on $\triangle ABD$ :

$BD^2 = 13^2 + 46^2 - 2 \cdot 13 \cdot 46 \cdot \cos 120^{\circ} = 2883$

So $AC = \frac{2}{\sqrt{3}} \cdot \sqrt{2883} = 62$ .

See also

1998 AHSME (Problems)
Preceded by Problem 25	Followed by Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1998_AHSME_Problems/Problem_26"

Category: Introductory Geometry Problems

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