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Difference between revisions of "1998 AHSME Problems/Problem 26"

(→‎Solution: 2nd <asy> image, this is kind've fun)
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== Problem ==
 
== Problem ==
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In quadrilateral <math>ABCD</math>, it is given that <math>\angle A = 120^{\circ}</math>, angles <math>B</math> and <math>D</math> are right angles, <math>AB = 13</math>, and <math>AD = 46</math>. Then <math>AC=</math>
+
In [[quadrilateral]] <math>ABCD</math>, it is given that <math>\angle A = 120^{\circ}</math>, angles <math>B</math> and <math>D</math> are [[right angle]]s, <math>AB = 13</math>, and <math>AD = 46</math>. Then <math>AC=</math>
 +
 
 
<math>\mathrm{(A)}\ 60
 
<math>\mathrm{(A)}\ 60
 
\qquad\mathrm{(B)}\ 62
 
\qquad\mathrm{(B)}\ 62
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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Let the extensions of <math>\overline{DA}</math> and <math>\overline{CB}</math> be at <math>E</math>. Since <math>\angle BAD = 120^{\circ}</math>, <math>\angle BAE = 60^{\circ}</math> and <math>\triangle ABE</math> is a <tt>30-60-90</tt> triangle. Also, <math>\triangle ABE \sim \triangle CDE</math>, so <math>\triangle CDE</math> is also a <tt>30-60-90</tt> triangle.
+
Let the extensions of <math>\overline{DA}</math> and <math>\overline{CB}</math> be at <math>E</math>. Since <math>\angle BAD = 120^{\circ}</math>, <math>\angle BAE = 60^{\circ}</math> and <math>\triangle ABE</math> is a <math>30-60-90</math> triangle. Also, <math>\triangle ABE \sim \triangle CDE</math>, so <math>\triangle CDE</math> is also a <math>30-60-90</math> triangle.
  
 
<center><asy>
 
<center><asy>
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</asy></center>
 
</asy></center>
  
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Opposite angles add up to <math>180^{\circ}</math>, so <math>ABCD</math> is a cyclic quadrilateral. Also, <math>\angle B = \angle D = 90^{\circ}</math>, from which it follows that <math>\overline{AC}</math> is a diameter of the circumscribing circle. We can apply the extended version of the [[Law of Sines]] on <math>\triangle ABD</math>:
+
Opposite angles add up to <math>180^{\circ}</math>, so <math>ABCD</math> is a [[cyclic quadrilateral]]. Also, <math>\angle B = \angle D = 90^{\circ}</math>, from which it follows that <math>\overline{AC}</math> is a diameter of the circumscribing circle. We can apply the extended version of the [[Law of Sines]] on <math>\triangle ABD</math>:
  
 
<cmath>AC = 2R = \frac{BD}{\sin 120^{\circ}} = \frac{2}{\sqrt{3}} BD</cmath>  
 
<cmath>AC = 2R = \frac{BD}{\sin 120^{\circ}} = \frac{2}{\sqrt{3}} BD</cmath>  

Revision as of 22:14, 9 February 2008

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 $AC=$

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

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.

[asy] 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, R=(A+E)/2, T=(A+B)/2; draw(D--A--B--C--cycle); draw(C--A); draw(A--E--B,dashed); label("\(A\)",A,SSW); label("\(B\)",B,NNE); label("\(C\)",C,WNW); label("\(D\)",D,SSW); label("\(E\)",E,SSE); label("24\(\sqrt{3}\)",P,W); label("46",Q,S); label("26",R,S); label("13",T,WNW); [/asy]


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

[asy] 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+A)/2, T=(A+B)/2; draw(D--A--B--C--cycle); draw(B--D,dashed); draw(A--C,dashed); draw(circumcircle(A,B,C)); label("\(A\)",A,SSW); label("\(B\)",B,NNE); label("\(C\)",C,WNW); label("\(D\)",D,SSW); label("\(O\)",circumcenter(A,B,C),SW); dot(circumcenter(A,B,C)); label("46",Q,S); label("13",T,E); [/asy]

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 • Answer Key • Resources)
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
All AHSME Problems and Solutions
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