Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 AIME Problems/Problem 9"
Line 69: Line 69:
  
 
The answer is <math>\boxed{616}</math>.
 
The answer is <math>\boxed{616}</math>.
 +
 +
== Solution 5 ==
 +
 +
Suppose <math>\angle BAM=\angle CAM =x</math>, since <math>\angle BDC=3\angle BAC</math>, we have <math>\angle BDM=\angle MDC = 3x</math>. Therefore, <math>\angle DBC=\angle DCB = 90^\circ -3x</math> and <math>\angle ABD=\angle DCA=2x</math>. As a result, <math>\triangle KAC</math> is isosceles, <math>KC=KA</math>.
 +
 +
Let <math>H</math> be a point on the extension of <math>CD</math> through <math>D</math> such that <math>\overline{HB}\perp\overline{BC}</math> and denote the intersection of <math>\overline{HC}</math> and <math>\overline{AB}</math> as <math>K</math>. Then, <math>BH=2DM=2, \overline{HB}\parallel\overline{DM}</math>, and <math>HD=DC</math> by the [[Midpoint]] Theorem. So, <math>\angle HBA=x</math> and <math>\angle CDM=\angle CHB=\angle HDA= 3x</math>.
 +
 +
Consequently, <math>\triangle HBK\sim \triangle DAK</math>, <cmath>\frac{BK}{KA}=\frac{HK}{KD}=\frac{1}{5}</cmath> Assume <math>BK=a</math> and <math>HK=b</math>, then <math>KA=5a</math> and <math>KD = 5b</math>. Since <math>KC=KA, KC=5a</math>, and since <math>HD=DC</math>, <math>KC=11b</math>. Therefore, <math>a=\frac{11}{5}b</math>.
 +
 +
In <math>\triangle BDM</math>, by the [[Pythagorean Theorem]], <math>BM=\sqrt{36b^2-1}</math>. Similarly in <math>\triangle BAM</math>, <math>BM=\sqrt{36a^2-121}</math>. So <cmath>\sqrt{36a^2-121}=\sqrt{36b^2-1}</cmath> Since <math>a=\frac{11}{5}b</math>, we have <math>b=\frac{5\sqrt{5}}{12}</math> and <math>a=\frac{11\sqrt{5}}{12}</math>. Consequently, <math>BM=\frac{11}{2}</math> and <math>AB=\frac{11\sqrt{5}}{2}</math>. Thus, the perimeter of <math>\triangle ABC</math> is <math>11+\sqrt{605}</math>, and the answer is <math>\boxed{616}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:47, 3 February 2023

Problem

Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$

[asy] import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(1,1.63),W); dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE); dot((1,0),ds); label("$M$",(1,0),S); dot((1,0.7),ds); label("$D$",(1,0.7),NE); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]

Solution 1

Let $x=\angle CAM$, so $3x=\angle CDM$. Then, $\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11$. Expanding $\tan 3x$ using the angle sum identity gives \[\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.\] Thus, $\frac{3-\tan^2x}{1-3\tan^2x}=11$. Solving, we get $\tan x= \frac 12$. Hence, $CM=\frac{11}2$ and $AC= \frac{11\sqrt{5}}2$ by the Pythagorean Theorem. The total perimeter is $2(AC + CM) = \sqrt{605}+11$. The answer is thus $a+b=\boxed{616}$.

Solution 2

In a similar fashion, we encode the angles as complex numbers, so if $BM=x$, then $\angle BAD=\text{Arg}(11+xi)$ and $\angle BDM=\text{Arg}(1+xi)$. So we need only find $x$ such that $\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)$. This will happen when $\frac{363x-x^3}{1331-33x^2}=x$, which simplifies to $121x-4x^3=0$. Therefore, $x=\frac{11}{2}$. By the Pythagorean Theorem, $AB=\frac{11\sqrt{5}}{2}$, so the perimeter is $11+11\sqrt{5}=11+\sqrt{605}$, giving us our answer, $\boxed{616}$.

Solution 3

Let $\angle BAD=\alpha$, so $\angle BDM=3\alpha$, $\angle BDA=180-3\alpha$, and thus $\angle ABD=2\alpha.$ We can then draw the angle bisector of $\angle ABD$, and let it intersect $\overline{AM}$ at $E.$ Since $\angle BAE=\angle ABE$, $AE=BE.$ Let $AE=x$. Then we see by the Pythagorean Theorem, $BM=\sqrt{BE^2-ME^2}=\sqrt{x^2-(11-x)^2}=\sqrt{22x-121}$, $BD=\sqrt{BM^2+1}=\sqrt{22x-120}$, $BA=\sqrt{BM^2+121}=\sqrt{22x}$, and $DE=10-x.$ By the angle bisector theorem, $BA/BD=EA/ED.$ Substituting in what we know for the lengths of those segments, we see that \[\frac{\sqrt{22x}}{\sqrt{22x-120}}=\frac{x}{10-x}.\] multiplying by both denominators and squaring both sides yields \[22x(10-x)^2=x^2(22x-120)\] which simplifies to $x=\frac{55}{8}.$ Substituting this in for x in the equations for $BA$ and $BM$ yields $BA=\frac{\sqrt{605}}{2}$ and $BM=\frac{11}{2}.$ Thus the perimeter is $11+\sqrt{605}$, and the answer is $\boxed{616}$.

Solution 4

The triangle is symmetrical so we can split it in half ($\triangle ABM$ and $\triangle ACM$).

Let $\angle BAM = y$ and $\angle BDM = 3y$. By the Law of Sines on triangle $BAD$, $\frac{10}{\sin 2y} = \frac{BD}{\sin y}$. Using $\sin 2y = 2\sin y\cos y$ we can get $BD = \frac{5}{\cos y}$. We can use this information to relate $BD$ to $DM$ by using the Law of Sines on triangle $BMD$.

\[\frac{\frac{5}{\cos y}}{\sin BMD} = \frac{1}{\sin 90^\circ - 3y}\]

$\sin BMD = 1$ (as $\angle BMD$ is a right angle), so $\frac{1}{\sin 90^\circ - 3y} = \frac{5}{\cos y}$. Using the identity $\sin 90^\circ - x = \cos x$, we can turn the equation into::

\[\frac{1}{\cos 3y} = \frac{5}{\cos y}\]

\[5\cos 3y = \cos y\]

\[5(4\cos ^3 y - 3\cos y) = \cos y\]

\[20\cos ^3 y = 16 \cos y\]

\[5\cos ^3 y = 4\cos y\]

\[5\cos ^2 y = 4\]

\[\cos ^2 y = \frac{4}{5}\]

Now that we've found $\cos y$, we can look at the side lengths of $BM$ and $AB$ (since they are symmetrical, the perimeter of $\triangle ABC$ is $2(BM + AB)$.

We note that $BM = 11\tan y$ and $AB = 11\sec y$.

\[\sin ^2 y = 1 - \cos ^2 y\]

\[\sin ^2 y = \frac{1}{5}\]

\[\tan ^2 y = \frac{1}{4}\]

\[\tan y = \frac{1}{2}\]

(Note it is positive since $BM > 0$).

\[\sec ^2 y = \frac{5}{4}\]

\[\sec y = \frac{\sqrt{5}}{2}\]

\[BM + AB = 11\frac{\sqrt{5}+1}{2}\]

\[2(BM + AB) = 11(\sqrt{5} + 1)\]

\[2(BM + AB) = 11\sqrt{5} + 11\]

\[2(BM + AB) = \sqrt{605} + 11\]

The answer is $\boxed{616}$.

Solution 5

Suppose $\angle BAM=\angle CAM =x$, since $\angle BDC=3\angle BAC$, we have $\angle BDM=\angle MDC = 3x$. Therefore, $\angle DBC=\angle DCB = 90^\circ -3x$ and $\angle ABD=\angle DCA=2x$. As a result, $\triangle KAC$ is isosceles, $KC=KA$.

Let $H$ be a point on the extension of $CD$ through $D$ such that $\overline{HB}\perp\overline{BC}$ and denote the intersection of $\overline{HC}$ and $\overline{AB}$ as $K$. Then, $BH=2DM=2, \overline{HB}\parallel\overline{DM}$, and $HD=DC$ by the Midpoint Theorem. So, $\angle HBA=x$ and $\angle CDM=\angle CHB=\angle HDA= 3x$.

Consequently, $\triangle HBK\sim \triangle DAK$, \[\frac{BK}{KA}=\frac{HK}{KD}=\frac{1}{5}\] Assume $BK=a$ and $HK=b$, then $KA=5a$ and $KD = 5b$. Since $KC=KA, KC=5a$, and since $HD=DC$, $KC=11b$. Therefore, $a=\frac{11}{5}b$.

In $\triangle BDM$, by the Pythagorean Theorem, $BM=\sqrt{36b^2-1}$. Similarly in $\triangle BAM$, $BM=\sqrt{36a^2-121}$. So \[\sqrt{36a^2-121}=\sqrt{36b^2-1}\] Since $a=\frac{11}{5}b$, we have $b=\frac{5\sqrt{5}}{12}$ and $a=\frac{11\sqrt{5}}{12}$. Consequently, $BM=\frac{11}{2}$ and $AB=\frac{11\sqrt{5}}{2}$. Thus, the perimeter of $\triangle ABC$ is $11+\sqrt{605}$, and the answer is $\boxed{616}$.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_9&oldid=188559"