Difference between revisions of "1998 AHSME Problems/Problem 26"
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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 \ | + | 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. |
| − | { | + | <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>, < | + | |
| + | 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 
, 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 30-60-90 triangle. Also, 
, so 
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 
, and 
. By the Pythagorean Theorem on 
, ![\[AC = \sqrt{(46)^2 + (24\sqrt{3})^2} = 62 \Rightarrow \mathrm{(B)}.\]](http://latex.artofproblemsolving.com/8/f/d/8fdcdf597041804dfeff9c3c580977e717c0779d.png)
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 
, 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 
:
![\[AC = 2R = \frac{BD}{\sin 120^{\circ}} = \frac{2}{\sqrt{3}} BD\]](http://latex.artofproblemsolving.com/4/c/a/4caa5fda3c0d53824d9b7aff5131dec8b05dd263.png)
By the Law of Cosines on :
![\[BD^2 = 13^2 + 46^2 - 2 \cdot 13 \cdot 46 \cdot \cos 120^{\circ} = 2883\]](http://latex.artofproblemsolving.com/4/a/9/4a9d839d943f73e8a78e7f9821c1749b7b72c54e.png)
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 | ||
