Difference between revisions of "2001 AIME II Problems/Problem 13"
(→Solution 3) |
Pi is 3.14 (talk | contribs) (→Video Solution) |
||
(One intermediate revision by one other user not shown) | |||
Line 34: | Line 34: | ||
~bluesoul | ~bluesoul | ||
− | == Video Solution == | + | == Solution 3 == |
+ | Denote <math>\angle{BAD}=\angle{CDA}=x</math>, and <math>\angle{ABD}=\angle{BCD}=y</math>. Note that <math>\angle{ADB}=180^\circ-x-y</math>, and <math>\angle{DBC}=360^\circ-2x-2y</math>. This motivates us to draw the angle bisector of <math>\angle{DBC}</math> because <math>\angle{DBC} = 2 \angle{ADB}</math>, so we do so and consider the intersection with <math>CD</math> as <math>E</math>. By the angle bisector theorem, we have <math>\frac{CE}{DE} = \frac{BC}{BD} = \frac{3}{5}</math>, so we write <math>CE=3z</math> and <math>DE=5z</math>. We also know that <math>\angle{EBC}=\angle{ADB}</math> and <math>\angle{BCE}=\angle{DBA}</math>, so <math>\triangle{ADB} \sim \triangle{EBC}</math>. Hence, <math>\frac{CE}{BC}=\frac{AB}{BD}</math>, so we have <math>3z=\frac{24}{5}</math>. As <math>CD=8z</math>, it must be that <math>CD=\frac{64}{5}</math>, so the final answer is <math>\boxed{069}</math>. | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
https://youtu.be/NsQbhYfGh1Q?t=75 | https://youtu.be/NsQbhYfGh1Q?t=75 | ||
Latest revision as of 03:32, 23 January 2023
Contents
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
Solution 3
Denote , and
. Note that
, and
. This motivates us to draw the angle bisector of
because
, so we do so and consider the intersection with
as
. By the angle bisector theorem, we have
, so we write
and
. We also know that
and
, so
. Hence,
, so we have
. As
, it must be that
, so the final answer is
.
Video Solution by OmegaLearn
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.