Difference between revisions of "2012 AIME I Problems/Problem 12"
Boxtheanswer (talk | contribs) m (→Solution 9) |
|||
(17 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | ==Problem== |
Let <math>\triangle ABC</math> be a right triangle with right angle at <math>C.</math> Let <math>D</math> and <math>E</math> be points on <math>\overline{AB}</math> with <math>D</math> between <math>A</math> and <math>E</math> such that <math>\overline{CD}</math> and <math>\overline{CE}</math> trisect <math>\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, and <math>p</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p.</math> | Let <math>\triangle ABC</math> be a right triangle with right angle at <math>C.</math> Let <math>D</math> and <math>E</math> be points on <math>\overline{AB}</math> with <math>D</math> between <math>A</math> and <math>E</math> such that <math>\overline{CD}</math> and <math>\overline{CE}</math> trisect <math>\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, and <math>p</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p.</math> | ||
− | |||
− | |||
− | |||
− | === Solution 2 | + | == Solution 1 == |
+ | We have <math>\angle BCE = \angle ECD = \angle DCA = \tfrac 13 \cdot 90^\circ = 30^\circ</math>. Drop the altitude from <math>D</math> to <math>CB</math> and call the foot <math>F</math>. | ||
+ | <center><asy>import cse5;size(200); | ||
+ | defaultpen(linewidth(0.4)+fontsize(8)); | ||
+ | |||
+ | pair A,B,C,D,E0,F; | ||
+ | C=origin; | ||
+ | B=(10,0); | ||
+ | A=(0,5); | ||
+ | E0=extension(C,dir(30),A,B); | ||
+ | D=extension(C,dir(60),A,B); | ||
+ | F=foot(D,C,B); | ||
+ | |||
+ | draw(A--B--C--A, black+0.8); | ||
+ | draw(C--D--F^^C--E0, gray); | ||
+ | dot("$A$",A,N); | ||
+ | dot("$B$",B,SE); | ||
+ | dot("$C$",C,SW); | ||
+ | dot("$D$",D,NE); | ||
+ | dot("$E$",E0,2*NE); | ||
+ | dot("$F$",F,S); | ||
+ | |||
+ | label("$8$",D--E0,2*NE); | ||
+ | label("$15$",E0--B,2*NE); | ||
+ | label("$11a$",C--B,2*S); | ||
+ | label(rotate(60)*"$8a$",C--D,2*NW); | ||
+ | label(rotate(-90)*"$4\sqrt{3}a$",D--F, E); | ||
+ | label("$4a$",C--F,2*S); | ||
+ | </asy></center> | ||
+ | Let <math>CD = 8a</math>. Using angle bisector theorem on <math>\triangle CDB</math>, we get <math>CB = 15a</math>. Now <math>CDF</math> is a <math>30</math>-<math>60</math>-<math>90</math> triangle, so <math>CF = 4a</math>, <math>FD = 4a\sqrt{3}</math>, and <math>FB = 11a</math>. Finally, <math>\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}</math>. Our final answer is <math>4 + 3 + 11 = \boxed{018}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
Without loss of generality, set <math>CB = 1</math>. Then, by the Angle Bisector Theorem on triangle <math>DCB</math>, we have <math>CD = \frac{8}{15}</math>. We apply the Law of Cosines to triangle <math>DCB</math> to get <math>1 + \frac{64}{225} - \frac{8}{15} = BD^{2}</math>, which we can simplify to get <math>BD = \frac{13}{15}</math>. | Without loss of generality, set <math>CB = 1</math>. Then, by the Angle Bisector Theorem on triangle <math>DCB</math>, we have <math>CD = \frac{8}{15}</math>. We apply the Law of Cosines to triangle <math>DCB</math> to get <math>1 + \frac{64}{225} - \frac{8}{15} = BD^{2}</math>, which we can simplify to get <math>BD = \frac{13}{15}</math>. | ||
Line 13: | Line 41: | ||
Our final answer is <math>4+3+11 = \boxed{018}</math>. | Our final answer is <math>4+3+11 = \boxed{018}</math>. | ||
− | + | == Solution 3 == | |
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig) | (This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig) | ||
Line 37: | Line 65: | ||
<math>\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}</math>. Our final answer is <math>4 + 3 + 11 = \boxed{018}</math>. | <math>\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}</math>. Our final answer is <math>4 + 3 + 11 = \boxed{018}</math>. | ||
− | + | == Solution 4 == | |
(This solution avoids advanced trigonometry) | (This solution avoids advanced trigonometry) | ||
Line 50: | Line 78: | ||
Thus <math>\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}</math>, and our answer is <math>4+3+11=\boxed{018}</math>. | Thus <math>\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}</math>, and our answer is <math>4+3+11=\boxed{018}</math>. | ||
− | + | == Solution 5 == | |
(Another solution without trigonometry) | (Another solution without trigonometry) | ||
Line 61: | Line 89: | ||
From <math>\triangle AFD \sim \triangle BCD</math>, we have that <math>\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}</math>. Simplifying yields <math>\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1</math>, and <math>\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}</math>, so our answer is <math>4+3+11=\boxed{018}</math>. | From <math>\triangle AFD \sim \triangle BCD</math>, we have that <math>\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}</math>. Simplifying yields <math>\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1</math>, and <math>\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}</math>, so our answer is <math>4+3+11=\boxed{018}</math>. | ||
− | + | == Solution 6 == | |
Let <math>CB = 1</math>, and let the feet of the altitudes from <math>D</math> and <math>E</math> to <math>\overline{CB}</math> be <math>D'</math> and <math>E'</math>, respectively. Also, let <math>DE = 8k</math> and <math>EB = 15k</math>. We see that <math>BD' = 15k\cos B</math> and <math>BE' = 23k\cos B</math> by right triangles <math>\triangle{BDD'}</math> and <math>\triangle{BEE'}</math>. From this we have that <math>D'E' = 8k\cos B</math>. With the same triangles we have <math>DD' = 23k\sin B</math> and <math>EE' = 15k\sin B</math>. From 30-60-90 triangles <math>\triangle{CDD'}</math> and <math>\triangle{CEE'}</math>, we see that <math>CD' = \frac{23k\sqrt{3}\sin B}{3}</math> and <math>CE' = 15k\sqrt{3}\sin B</math>, so <math>D'E' = \frac{22k\sqrt{3}\sin B}{3}</math>. From our two values of <math>D'E'</math> we get: | Let <math>CB = 1</math>, and let the feet of the altitudes from <math>D</math> and <math>E</math> to <math>\overline{CB}</math> be <math>D'</math> and <math>E'</math>, respectively. Also, let <math>DE = 8k</math> and <math>EB = 15k</math>. We see that <math>BD' = 15k\cos B</math> and <math>BE' = 23k\cos B</math> by right triangles <math>\triangle{BDD'}</math> and <math>\triangle{BEE'}</math>. From this we have that <math>D'E' = 8k\cos B</math>. With the same triangles we have <math>DD' = 23k\sin B</math> and <math>EE' = 15k\sin B</math>. From 30-60-90 triangles <math>\triangle{CDD'}</math> and <math>\triangle{CEE'}</math>, we see that <math>CD' = \frac{23k\sqrt{3}\sin B}{3}</math> and <math>CE' = 15k\sqrt{3}\sin B</math>, so <math>D'E' = \frac{22k\sqrt{3}\sin B}{3}</math>. From our two values of <math>D'E'</math> we get: | ||
<math>8k\cos B = \frac{22k\sqrt{3}\sin B}{3}</math> | <math>8k\cos B = \frac{22k\sqrt{3}\sin B}{3}</math> | ||
Line 70: | Line 98: | ||
Our answer is then <math>4+3+11 = \boxed{018}</math>. | Our answer is then <math>4+3+11 = \boxed{018}</math>. | ||
− | + | == Video Solution by Richard Rusczyk == | |
https://artofproblemsolving.com/videos/amc/2012aimei/352 | https://artofproblemsolving.com/videos/amc/2012aimei/352 | ||
− | ~ | + | ==Solution 7== |
+ | WLOG, let <math>DE=8</math> and <math>BE=15</math>. First, by the Law of Sines on <math>\triangle CEB</math>, we find that | ||
+ | <cmath>\frac{\sin\angle B}{CE}=\frac{\sin\angle ECB}{BE}=\frac{1}{30}\implies \sin\angle B=\frac{CE}{30}.</cmath> | ||
+ | Now, we will find <math>CE</math>. Consider the following diagram: | ||
+ | [[Image:2012 AIME I -12 Sol 7 Diagram.png|200px|thumb|left| ]] | ||
+ | We have constructed equilateral triangle <math>\triangle BDP</math>, and its circumcircle. Since <math>\angle DCB=\angle DPB=60^\circ</math>, <math>C</math> lies on <math>(BDP)</math> as well. Let <math>Q</math> be the point diametrically opposite <math>P</math> on <math>(BCD)</math>, and let <math>R</math> be the foot of <math>Q</math> on <math>BD</math> (this is the midpoint of <math>BD</math>). It is easy to compute that <math>RQ=\frac{23}{2\sqrt3}</math> and <math>ER=\frac{23}{2}-8=\frac{7}{2}</math>. Therefore, by the Pythagorean Theorem, <math>EQ=\frac{13}{\sqrt{3}}</math>. Now, by Power of a Point, we know that <math>(DE)(BE)=(EQ)(EC)</math>, which means that | ||
+ | <cmath>120=\frac{13EC}{\sqrt{3}}\implies EC=\frac{120\sqrt{3}}{13}.</cmath> | ||
+ | From before, we know that <math>\sin\angle B=\frac{EC}{30}\implies \sin\angle B=\frac{4\sqrt3}{13}</math>. It's now easy to compute <math>\cos\angle B</math> as well using the Pythagorean identity; we find that <math>\cos\angle B=\frac{11}{13}</math>, and thus <math>\tan\angle B=\frac{4\sqrt3}{11}</math> for an answer of <math>\boxed{018}</math>. | ||
+ | -brainiacmaniac31 | ||
+ | |||
+ | ==Solution 8== | ||
+ | [[File:2012 AIME 12.png|500px]] | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 9== | ||
+ | WLOG, let <math>DE=8</math> and <math>EB=15</math>. (this will be redefined later) | ||
+ | Define points <math>A'</math>, <math>D'</math>, <math>E'</math>, and <math>B'</math> such that <math>A'</math> is on <math>AC</math>, <math>B'</math> is on <math>CB</math> and <math>D'</math> and <math>E'</math> are the intersections of <math>A'B'</math> with <math>CD</math> and <math>CE</math>, with <math>CD' = CE' = 1</math>, respectively. From cross ratios, we have: | ||
+ | \begin {align*} | ||
+ | \frac{(AE)(DB)}{(AB)(DE)} &= \frac{(A'E')(D'B')}{(A'B')(D'E')} \\ | ||
+ | \frac{(AD+8)(23)}{(AD+23)(8)} & = \frac{(\cos(15)+\sin(15))^2}{(2 \cos(15))(2 \sin(15))} \\ | ||
+ | & \implies AD = 92/11 | ||
+ | \end {align*} | ||
+ | For simplicity, scale everything by <math>11</math>, so <math>AD=92</math>, <math>DE=88</math> and <math>EB = 165</math>. | ||
+ | From the ratio lemma, we have: | ||
+ | \begin {align*} | ||
+ | \frac{AC}{CB} &= \frac{AD \sin{\angle BCD}}{DB \sin{\angle ACD}} \\ | ||
+ | \tan B &= \frac{92 \cdot \sqrt{3}/2}{253 \cdot 1/2} \\ | ||
+ | \tan B &= \frac{4 \sqrt{3}}{2}\\ | ||
+ | &\implies \boxed{018}. | ||
+ | \end{align*} | ||
+ | ~ boxtheanswer | ||
== See also == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=11|num-a=13}} | {{AIME box|year=2012|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:26, 12 September 2024
Contents
Problem
Let be a right triangle with right angle at Let and be points on with between and such that and trisect If then can be written as where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. Find
Solution 1
We have . Drop the altitude from to and call the foot .
Let . Using angle bisector theorem on , we get . Now is a -- triangle, so , , and . Finally, . Our final answer is .
Solution 2
Without loss of generality, set . Then, by the Angle Bisector Theorem on triangle , we have . We apply the Law of Cosines to triangle to get , which we can simplify to get .
Now, we have by another application of the Law of Cosines to triangle , so . In addition, , so .
Our final answer is .
Solution 3
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)
Find values for all angles in terms of . , , , , and .
Use the law of sines on and :
In , . This simplifies to .
In , . This simplifies to .
Solve for and equate them so that you get .
From this, .
Use a trig identity on the denominator on the right to obtain:
This simplifies to
This gives Dividing by , we have
. Our final answer is .
Solution 4
(This solution avoids advanced trigonometry)
Let be the foot of the perpendicular from to , and let be the foot of the perpendicular from to .
Now let . Clearly, triangles and are similar with , so .
Since triangles and are 30-60-90 right triangles, we can easily find other lengths in terms of . For example, we see that and . Therefore .
Again using the fact that triangles and are similar, we see that , so .
Thus , and our answer is .
Solution 5
(Another solution without trigonometry)
Extend to point such that . It is then clear that is similar to .
Let , . Then .
With the Angle Bisector Theorem, we get that . From 30-60-90 , we get that and .
From , we have that . Simplifying yields , and , so our answer is .
Solution 6
Let , and let the feet of the altitudes from and to be and , respectively. Also, let and . We see that and by right triangles and . From this we have that . With the same triangles we have and . From 30-60-90 triangles and , we see that and , so . From our two values of we get:
Our answer is then .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/352
Solution 7
WLOG, let and . First, by the Law of Sines on , we find that Now, we will find . Consider the following diagram:
We have constructed equilateral triangle , and its circumcircle. Since , lies on as well. Let be the point diametrically opposite on , and let be the foot of on (this is the midpoint of ). It is easy to compute that and . Therefore, by the Pythagorean Theorem, . Now, by Power of a Point, we know that , which means that From before, we know that . It's now easy to compute as well using the Pythagorean identity; we find that , and thus for an answer of . -brainiacmaniac31
Solution 8
vladimir.shelomovskii@gmail.com, vvsss
Solution 9
WLOG, let and . (this will be redefined later) Define points , , , and such that is on , is on and and are the intersections of with and , with , respectively. From cross ratios, we have: \begin {align*} \frac{(AE)(DB)}{(AB)(DE)} &= \frac{(A'E')(D'B')}{(A'B')(D'E')} \\ \frac{(AD+8)(23)}{(AD+23)(8)} & = \frac{(\cos(15)+\sin(15))^2}{(2 \cos(15))(2 \sin(15))} \\ & \implies AD = 92/11 \end {align*} For simplicity, scale everything by , so , and . From the ratio lemma, we have: \begin {align*} \frac{AC}{CB} &= \frac{AD \sin{\angle BCD}}{DB \sin{\angle ACD}} \\ \tan B &= \frac{92 \cdot \sqrt{3}/2}{253 \cdot 1/2} \\ \tan B &= \frac{4 \sqrt{3}}{2}\\ &\implies \boxed{018}. \end{align*} ~ boxtheanswer
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.