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Difference between revisions of "2001 AIME II Problems/Problem 13"

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== Solution 2 ==
 
== Solution 2 ==
  
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Draw a line from <math>B</math>, parallel to <math>\overline{AD}</math>, and let it meet <math>\overline{CD}</math> at <math>M</math>. Note that <math>\triangle{DAB}</math> is similar to <math>\triangle{BMC}</math> by AA similarity, since <math>\angle{ABD}=\angle{MCB}</math> and since <math>BM</math> is parallel to <math>CD</math> then <math>\angle{BMC}=\angle{ADM}=\angle{DAB}</math>. Now since <math>ADMB</math> is a trapezoid, <math>MD=8</math>. By the similarity, we have <math>MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}</math>, hence <math>CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}</math>.
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Draw a line from <math>B</math>, parallel to <math>\overline{AD}</math>, and let it meet <math>\overline{CD}</math> at <math>M</math>. Note that <math>\triangle{DAB}</math> is similar to <math>\triangle{BMC}</math> by AA similarity, since <math>\angle{ABD}=\angle{MCB}</math> and since <math>BM</math> is parallel to <math>CD</math> then <math>\angle{BMC}=\angle{ADM}=\angle{DAB}</math>. Now since <math>ADMB</math> is an isosceles trapezoid, <math>MD=8</math>. By the similarity, we have <math>MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}</math>, hence <math>CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:16, 7 January 2017

Problem

In quadrilateral $ABCD$, $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$, $AB = 8$, $BD = 10$, and $BC = 6$. The length $CD$ may be written in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution 1

Extend $\overline{AD}$ and $\overline{BC}$ to meet at $E$. Then, since $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCE$, we know that $\triangle ABD \sim \triangle DCE$. Hence $\angle ADB = \angle DEC$, and $\triangle BDE$ is isosceles. Then $BD = BE = 10$.

[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]

Using the similarity, we have:

\[\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5\]

The answer is $m+n = \boxed{069}$.


Extension: To Find $AD$, use Law of Cosines on $\triangle BCD$ to get $\cos(\angle BCD)=\frac{13}{20}$ Then since $\angle BCD=\angle ABD$ use Law of Cosines on $\triangle ABD$ to find $AD=2\sqrt{15}$

Solution 2

Draw a line from $B$, parallel to $\overline{AD}$, and let it meet $\overline{CD}$ at $M$. Note that $\triangle{DAB}$ is similar to $\triangle{BMC}$ by AA similarity, since $\angle{ABD}=\angle{MCB}$ and since $BM$ is parallel to $CD$ then $\angle{BMC}=\angle{ADM}=\angle{DAB}$. Now since $ADMB$ is an isosceles trapezoid, $MD=8$. By the similarity, we have $MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}$, hence $CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}$.

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

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