Difference between revisions of "2001 AIME II Problems/Problem 13"
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== Problem == | == Problem == | ||
+ | In [[quadrilateral]] <math>ABCD</math>, <math>\angle{BAD}\cong\angle{ADC}</math> and <math>\angle{ABD}\cong\angle{BCD}</math>, <math>AB = 8</math>, <math>BD = 10</math>, and <math>BC = 6</math>. The length <math>CD</math> may be written in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
− | == Solution == | + | == Solution 1 == |
+ | Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at <math>E</math>. Then, since <math>\angle BAD = \angle ADC</math> and <math>\angle ABD = \angle DCE</math>, we know that <math>\triangle ABD \sim \triangle DCE</math>. Hence <math>\angle ADB = \angle DEC</math>, and <math>\triangle BDE</math> is [[isosceles triangle|isosceles]]. Then <math>BD = BE = 10</math>. | ||
+ | |||
+ | <center><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></center> | ||
+ | |||
+ | Using the similarity, we have: | ||
+ | |||
+ | <cmath>\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5</cmath> | ||
+ | |||
+ | The answer is <math>m+n = \boxed{069}</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> | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | 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>. | ||
+ | == 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{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 | ||
+ | |||
+ | == 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 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2001|n=II|num-b=12|num-a=14}} | |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
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 |
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