2002 AMC 12B Problems/Problem 23

Problem

In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$ ?

$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$

Solution

Solution 1

Let $D$ be the foot of the median from $A$ to $\overline{BC}$ , and we let $AD = BC = 2a$ . Then by the Law of Cosines on $\triangle ABD, \triangle ACD$ , we have $\begin{align*} 1^2 &= a^2 + (2a)^2 - 2(a)(2a)\cos ADB \\ 2^2 &= a^2 + (2a)^2 - 2(a)(2a)\cos ADC \end{align*}$

Since $\cos ADC = \cos (180 - ADB) = -\cos ADB$ , we can add these two equations and get

$5 = 10a^2$

Hence $a = \frac{1}{\sqrt{2}}$ and $BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}$ .

Solution 2

From Stewart's Theorem, we have $(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.$ Simplifying, we get $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.$

Solution 3


unitsize(4cm);

pair A, B, C, D, M;

A = (1.768,0.935);
B = (1.414,0);
C = (0,0);
D = (1.768,0);
M = (0.707,0);

draw(A--B--C--cycle);
draw(A--D);
draw(D--B);
draw(A--M);

label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,S);
label("$M$",M,S)
label("$x$",(A+D)/2,E);
label("$y$",(B+D)/2,S);
label("$a$",(C+M)/2,S);
label("$a$",(M+B)/2,S);
label("$2a$",(A+M)/2,SE);
label("$1$",(A+B)/2,SE);
label("$2$",(A+C)/2,NW);

draw(rightanglemark(B,D,A,3));

 (Error making remote request. Unknown error_msg)

Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$ extended past $B$ . Let $AD = x$ and $BD = y$ . Using the Pythagorean Theorem, we obtain the equations

$\begin{align*} x^2 + y^2 = 1 \hspace{0.5cm}(1)\\ x^2 + y^2 + 2ya + a^2 = 4a^2 \hspace{0.5cm}(2)\\ x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3) \end{align*}$

Subtracting $(1)$ equation from $(2)$ and $(3)$ , we get

$\begin{align*} 2ya + a^2 = 4a^2 - 1 \hspace{0.5cm}(4)\\ 4ya + 4a^2 = 3 \hspace{0.5cm}(5) \end{align*}$

Then, subtracting $2 \times (4)$ from $(5)$ and rearranging, we get $10a^2 = 5$ , so $BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}$

~greenturtle 11/26/2017

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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