Difference between revisions of "1998 AHSME Problems/Problem 26"
m (→Solution 1: fix <asy>) |
(→Solution: 2nd <asy> image, this is kind've fun) |
||
Line 35: | Line 35: | ||
=== Solution 2 === | === Solution 2 === | ||
− | + | <center><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></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>: | ||
Line 44: | Line 63: | ||
<cmath>BD^2 = 13^2 + 46^2 - 2 \cdot 13 \cdot 46 \cdot \cos 120^{\circ} = 2883</cmath> | <cmath>BD^2 = 13^2 + 46^2 - 2 \cdot 13 \cdot 46 \cdot \cos 120^{\circ} = 2883</cmath> | ||
− | So <math>AC = \frac{2}{\sqrt{3}} \sqrt{2883} = 62</math>. | + | So <math>AC = \frac{2}{\sqrt{3}} \cdot \sqrt{2883} = 62</math>. |
== See also == | == See also == |
Revision as of 22:12, 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.
![[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 |