Difference between revisions of "2001 AIME II Problems/Problem 13"
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== Problem == | == Problem == | ||
In [[quadrilateral]] <math>ABCD</math>, <math>\angle{BAD}\cong\angle{ADC}</math> and <math>\angle{ABD}\cong\angle{BCD}</math>, <math>AB = 8</math>, <math>BD = 10</math>, and <math>BC = 6</math>. The length <math>CD</math> may be written in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | In [[quadrilateral]] <math>ABCD</math>, <math>\angle{BAD}\cong\angle{ADC}</math> and <math>\angle{ABD}\cong\angle{BCD}</math>, <math>AB = 8</math>, <math>BD = 10</math>, and <math>BC = 6</math>. The length <math>CD</math> may be written in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
+ | |||
+ | == Solution == | ||
+ | Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at <math>E</math>. Then, since <math>\angle BAD = \angle ADC</math> and <math>\angle ABD = \angle DCE</math>, we know that <math>\triangle ABD \sim \triangle DCE</math>. Hence <math>\angle ADB = \angle DEC</math>, and <math>\triangle BDE</math> is [[isosceles triangle|isosceles]]. Then <math>BD = BE = 10</math>. | ||
+ | |||
+ | <center><asy> | ||
+ | /* We arbitrarily set AD = x */ | ||
+ | real x = 60^.5, anglesize = 28; | ||
+ | |||
+ | pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7); | ||
+ | pair A=(0,0), D=(x,0), B=IP(CR(A,8),CR(D,10)), E=(-3x/5,0), C=IP(CR(E,16),CR(D,64/5)); | ||
+ | D(MP("A",A)--MP("B",B,NW)--MP("C",C,NW)--MP("D",D)--cycle); D(B--D); D(A--MP("E",E)--B,d); | ||
+ | D(anglemark(D,A,B,anglesize));D(anglemark(C,D,A,anglesize));D(anglemark(A,B,D,anglesize));D(anglemark(E,C,D,anglesize));D(anglemark(A,B,D,5/4*anglesize));D(anglemark(E,C,D,5/4*anglesize)); | ||
+ | MP("10",(B+D)/2,SW);MP("8",(A+B)/2,W);MP("6",(B+C)/2,NW); | ||
+ | </asy></center> | ||
+ | |||
+ | Using the similarity, we have: | ||
+ | |||
+ | <cmath>\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5</cmath> | ||
+ | |||
+ | The answer is <math>m+n = \boxed{069}</math>. | ||
+ | |||
+ | |||
+ | '''Extension''': Find <math>AD</math>. | ||
== See also == | == See also == |
Revision as of 13:18, 17 May 2014
Problem
In quadrilateral , and , , , and . The length may be written in the form , where and are relatively prime positive integers. Find .
Solution
Extend and to meet at . Then, since and , we know that . Hence , and is isosceles. Then .
Using the similarity, we have:
The answer is .
Extension: Find .
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.