1995 AIME Problems/Problem 9

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 $a+b.$

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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 $\boxed{616}$ .

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
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