Difference between revisions of "2001 AIME II Problems/Problem 13"
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== Solution 3 == | == Solution 3 == | ||
Since <math>\angle{BAD}=\angle{ADM}</math>, if we extend AB and DC, they must meet at one point to form a isosceles triangle <math>\triangle{ADM}</math>.Now, since the problem told that <math>\angle{ABD}=\angle{BCD}</math>, we can imply that <math>\angle{DBM}=\angle{BCM}</math> | Since <math>\angle{BAD}=\angle{ADM}</math>, if we extend AB and DC, they must meet at one point to form a isosceles triangle <math>\triangle{ADM}</math>.Now, since the problem told that <math>\angle{ABD}=\angle{BCD}</math>, we can imply that <math>\angle{DBM}=\angle{BCM}</math> | ||
− | Since <math>\angle{M}=\angle{M}</math>, so <math>\triangle{CBM}\sim\triangle{BDM}</math>. Assume the length of <math>BM=x</math>;Since <math>\frac{BC}{MB}=\frac{DB}{MD}</math> we can get <math>\frac{6}{x}=\frac{10}{8+x}</math>, we get that <math>x=12</math>. similarly, we use the same pair of similar triangle we get <math>\frac{CM}{BM}=\frac{BM}{DM}</math>, we get that <math>CM=\frac{36}{5}</math>. Finally, <math>CD=MD-MC=\frac{64}{5}\implies 64+ | + | Since <math>\angle{M}=\angle{M}</math>, so <math>\triangle{CBM}\sim\triangle{BDM}</math>. Assume the length of <math>BM=x</math>;Since <math>\frac{BC}{MB}=\frac{DB}{MD}</math> we can get <math>\frac{6}{x}=\frac{10}{8+x}</math>, we get that <math>x=12</math>.So <math>AM=DM=20</math> similarly, we use the same pair of similar triangle we get <math>\frac{CM}{BM}=\frac{BM}{DM}</math>, we get that <math>CM=\frac{36}{5}</math>. Finally, <math>CD=MD-MC=\frac{64}{5}\implies 64+5=69=\boxed{069}</math> |
~bluesoul | ~bluesoul | ||
+ | |||
== Video Solution == | == Video Solution == | ||
https://youtu.be/NsQbhYfGh1Q?t=75 | https://youtu.be/NsQbhYfGh1Q?t=75 |
Revision as of 22:44, 29 October 2021
Problem
In quadrilateral ,
and
,
,
, and
. The length
may be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
Extend and
to meet at
. Then, since
and
, we know that
. Hence
, and
is isosceles. Then
.
![[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]](http://latex.artofproblemsolving.com/4/1/7/417316517e7273e0d7f31dbbae8e0b80e30a8d54.png)
Using the similarity, we have:
The answer is .
Extension: To Find , use Law of Cosines on
to get
Then since
use Law of Cosines on
to find
Solution 2
Draw a line from , parallel to
, and let it meet
at
. Note that
is similar to
by AA similarity, since
and since
is parallel to
then
. Now since
is an isosceles trapezoid,
. By the similarity, we have
, hence
.
Solution 3
Since , if we extend AB and DC, they must meet at one point to form a isosceles triangle
.Now, since the problem told that
, we can imply that
Since
, so
. Assume the length of
;Since
we can get
, we get that
.So
similarly, we use the same pair of similar triangle we get
, we get that
. Finally,
~bluesoul
Video Solution
https://youtu.be/NsQbhYfGh1Q?t=75
~ pi_is_3.14
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.