Difference between revisions of "2001 AIME II Problems/Problem 13"
Pi is 3.14 (talk | contribs) (→Video Solution) |
|||
Line 37: | Line 37: | ||
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>. | 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 == | + | == Video Solution by OmegaLearn == |
https://youtu.be/NsQbhYfGh1Q?t=75 | https://youtu.be/NsQbhYfGh1Q?t=75 | ||
Latest revision as of 02:32, 23 January 2023
Contents
[hide]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 .
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.