Difference between revisions of "1998 AHSME Problems/Problem 26"
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== Problem == | == Problem == | ||
− | 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 | + | 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 === | ||
− | 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 < | + | 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> | ||
− | 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 , it is given that
, angles
and
are right angles,
, and
. Then
Solution
Solution 1
Let the extensions of and
be at
. Since
,
and
is a
triangle. Also,
, so
is also a
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]](http://latex.artofproblemsolving.com/c/6/e/c6e7459e1ed61b8bfc4ce08a34ee440ab75a6fbf.png)
Thus , and
. By the Pythagorean Theorem on
,
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]](http://latex.artofproblemsolving.com/6/8/e/68e66bc7b11557151e224b049d848d13c62ea5e2.png)
Opposite angles add up to , so
is a cyclic quadrilateral. Also,
, from which it follows that
is a diameter of the circumscribing circle. We can apply the extended version of the Law of Sines on
:
By the Law of Cosines on :
So .
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 |