Difference between revisions of "2002 AMC 12B Problems/Problem 23"

Revision as of 23:36, 26 November 2017

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

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

@@ Line 38: / Line 38: @@
 <cmath>
 \begin{align*}
-  x^2 + y^2 = 1 \t(1)\\
+  x^2 + y^2 = 1 \hspace{0.5cm}(1)\\
-x^2 + y^2 + 2ya + a^2 = 4a^2 (2)\\
+x^2 + y^2 + 2ya + a^2 = 4a^2 \hspace{0.5cm}(2)\\
-x^2 + y^2 + 4ya + 4a^2 = 4 (3)
+x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3)
 \end{align*}
 </cmath>
-Subtracting (1) equation from (2) and (3), we get
+Subtracting <math>(1)</math> equation from <math>(2)</math> and <math>(3)</math>, we get
 <cmath>
 \begin{align*}
-ya + a^2 = 4a^2 - 1 (4)\\
+ya + a^2 = 4a^2 - 1 \hspace{0.5cm}(4)\\
-ya + 4a^2 = 3 (5)
+ya + 4a^2 = 3 \hspace{0.5cm}(5)
 \end{align*}
 </cmath>
-Then, subtracting two times (4) from (5) and rearranging, we get <math>10a^2 = 5</math>, so <math>BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}</math>
+Then, subtracting <math>2 \times (4)</math> from <math>(5)</math> and rearranging, we get <math>10a^2 = 5</math>, so <math>BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}</math>
 ~greenturtle 11/26/2017

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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