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

(Undo revision 61762 by XXQw3rtyXx (talk))
Line 23: Line 23:
'''Extension''': Find <math>AD</math>.
'''Extension''': To Find <math>AD</math>, use Law of Cosines on <math>\triangle BCD</math> to get <math>\cos(\angle BCD)=\frac{13}{20}</math>
Then since <math>\angle BCD=\angle ABD</math> use Law of Cosines on <math>\triangle ABD</math> to find <math>AD=2\sqrt{15}</math>
== See also ==
== See also ==

Revision as of 16:05, 19 January 2016


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 $m + n$.


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 $AD=2\sqrt{15}$

See also

2001 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS