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

(→‎Solution 3)
(→‎Solution 3)
Line 32: Line 32:
  
 
=== Solution 3 ===
 
=== Solution 3 ===
 +
 +
<asy>
 +
 +
unitsize(4cm);
 +
 +
pair A, B, C, D, M;
 +
 +
A = (1.768,0.935);
 +
B = (1.414,0);
 +
C = (0,0);
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D = (1.768,0);
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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("$x$",(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));
 +
 +
</asy>
  
 
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>.
 
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>.

Revision as of 20:24, 27 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

[asy] 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("$x$",(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)); [/asy]

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

See also

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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