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1998 AHSME Problems/Problem 26: Difference between revisions

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solutions, {{image}} needed
 
→Solution 1: + <asy> image.
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== Solution ==
== Solution ==
=== Solution 1 ===
=== Solution 1 ===
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 \similar \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 <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.


{{image}}
<center><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<math>\sqrt{3}</math>",P,W);
label("46",Q,S);
label("26",R,S);
label("13",T,WNW);
</asy></center>


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


=== Solution 2 ===
=== Solution 2 ===

Revision as of 20:41, 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. 

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<math>\sqrt{3}</math>",P,W);
label("46",Q,S);
label("26",R,S);
label("13",T,WNW);
 (Error making remote request. Unknown error_msg)


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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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

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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