Difference between revisions of "2012 AIME I Problems/Problem 12"
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− | ==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 == | + | |
− | === Solution | + | == Solution 1 == |
+ | We have <math>\angle BCE = \angle ECD = \angle ECA = \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>. | ||
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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 28: | Line 58: | ||
Use a trig identity on the denominator on the right to obtain: <math>\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}</math> | Use a trig identity on the denominator on the right to obtain: <math>\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}</math> | ||
− | This simplifies to <math>\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \ | + | This simplifies to <math>\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}</math> |
This gives <math>8\sqrt{3}\cos B+8\sin B=30\sin B</math> | This gives <math>8\sqrt{3}\cos B+8\sin B=30\sin B</math> | ||
Line 35: | 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 48: | 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 58: | Line 88: | ||
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: | ||
+ | <math>8k\cos B = \frac{22k\sqrt{3}\sin B}{3}</math> | ||
+ | |||
+ | <math>\frac{\sin B}{\cos B} = \frac{8k}{\frac{22k\sqrt{3}}{3}} = \tan B</math> | ||
+ | |||
+ | <math>\tan B = \frac{8}{\frac{22\sqrt{3}}{3}} = \frac{24}{22\sqrt{3}} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}</math> | ||
+ | Our answer is then <math>4+3+11 = \boxed{018}</math>. | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | 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''' | ||
== 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 12:23, 7 September 2022
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
.
![[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]](http://latex.artofproblemsolving.com/9/d/4/9d434997e7b736adb2b4248a976dc241884d063a.png)
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
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.