2002 AMC 12B Problems/Problem 23: Difference between revisions

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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 $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 8: / Line 8: @@
 \qquad\mathrm{(E)}\ \sqrt{3}</math>
 == Solution ==
+[[Image:2002_12B_AMC-23.png]]
 === Solution 1 ===
-[[Image:2002_12B_AMC-23.png]]
 Let <math>D</math> be the foot of the median from <math>A</math> to <math>\overline{BC}</math>, and we let <math>AD = BC = 2a</math>. Then by the [[Law of Cosines]] on <math>\triangle ABD, \triangle ACD</math>, we have

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