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

(Solution 3)
(Solution 3)
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=== Solution 3 ===
 
=== Solution 3 ===
  
Let <math>D</math> be the foot of the altitude from <math>A</math> to <math>\overline{BC}</math> extended past <math>B</math>. Let <math>\overline{AD} = x</math> and <math>\overline{BD} = y</math>.
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Let <math>D</math> be the foot of the altitude from <math>A</math> to <math>\overline{BC}</math> extended past <math>B</math>. Let <math>AD = x</math> and <math>BD = y</math>.
 
Using the Pythagorean Theorem, we obtain the equations
 
Using the Pythagorean Theorem, we obtain the equations
  
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</cmath>
 
</cmath>
  
Then, subtracting two times the first equation from the second and rearranging, we get <math>2a^2 = 1</math>, so <math>2a = \sqrt{2}</math>
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Then, subtracting two times the first equation from the second and rearranging, we get <math>2a^2 = 1</math>, so <math>BC = 2a = \sqrt{2}</math>
  
 
== See also ==
 
== See also ==

Revision as of 23:12, 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

2002 12B AMC-23.png

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 \\ x^2 + y^2 + 2ya + a^2 = 4a^2 \\ x^2 + y^2 + 4ya + 4a^2 = 4 \end{align*}

Subtracting the first equation from the second and third equations, we get

\begin{align*} 2ya + a^2 = 4a^2 - 1 \\ 4ya + 4a^2 = 3 \end{align*}

Then, subtracting two times the first equation from the second and rearranging, we get $2a^2 = 1$, so $BC = 2a = \sqrt{2}$

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
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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