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

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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>.
 
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>.
  
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== Solution ==
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== 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>.  
 
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>.  
  
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'''Extension''': Find <math>AD</math>.  
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'''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>
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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>
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== Solution 2 ==
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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>.
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== Solution 3 ==
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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>
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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>
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~bluesoul
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== Solution 3 ==
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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>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/NsQbhYfGh1Q?t=75
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~ pi_is_3.14
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 02:32, 23 January 2023

Problem

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 $m + n$.

Solution 1

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 $AD=2\sqrt{15}$

Solution 2

Draw a line from $B$, parallel to $\overline{AD}$, and let it meet $\overline{CD}$ at $M$. Note that $\triangle{DAB}$ is similar to $\triangle{BMC}$ by AA similarity, since $\angle{ABD}=\angle{MCB}$ and since $BM$ is parallel to $CD$ then $\angle{BMC}=\angle{ADM}=\angle{DAB}$. Now since $ADMB$ is an isosceles trapezoid, $MD=8$. By the similarity, we have $MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}$, hence $CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}$.

Solution 3

Since $\angle{BAD}=\angle{ADM}$, if we extend AB and DC, they must meet at one point to form a isosceles triangle $\triangle{ADM}$.Now, since the problem told that $\angle{ABD}=\angle{BCD}$, we can imply that $\angle{DBM}=\angle{BCM}$ Since $\angle{M}=\angle{M}$, so $\triangle{CBM}\sim\triangle{BDM}$. Assume the length of $BM=x$;Since $\frac{BC}{MB}=\frac{DB}{MD}$ we can get $\frac{6}{x}=\frac{10}{8+x}$, we get that $x=12$.So $AM=DM=20$ similarly, we use the same pair of similar triangle we get $\frac{CM}{BM}=\frac{BM}{DM}$, we get that $CM=\frac{36}{5}$. Finally, $CD=MD-MC=\frac{64}{5}\implies 64+5=69=\boxed{069}$ ~bluesoul

Solution 3

Denote $\angle{BAD}=\angle{CDA}=x$, and $\angle{ABD}=\angle{BCD}=y$. Note that $\angle{ADB}=180^\circ-x-y$, and $\angle{DBC}=360^\circ-2x-2y$. This motivates us to draw the angle bisector of $\angle{DBC}$ because $\angle{DBC} = 2 \angle{ADB}$, so we do so and consider the intersection with $CD$ as $E$. By the angle bisector theorem, we have $\frac{CE}{DE} = \frac{BC}{BD} = \frac{3}{5}$, so we write $CE=3z$ and $DE=5z$. We also know that $\angle{EBC}=\angle{ADB}$ and $\angle{BCE}=\angle{DBA}$, so $\triangle{ADB} \sim \triangle{EBC}$. Hence, $\frac{CE}{BC}=\frac{AB}{BD}$, so we have $3z=\frac{24}{5}$. As $CD=8z$, it must be that $CD=\frac{64}{5}$, so the final answer is $\boxed{069}$.

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
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