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

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=== Solution 1 ===
 
=== Solution 1 ===
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[[Image:2002_12B_AMC-23.png]]
 
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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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<cmath>
 
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\begin{align*}
 
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1^2 &= a^2 + (2a)^2 - 2(a)(2a)\cos ADB \\
 
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2^2 &= a^2 + (2a)^2 - 2(a)(2a)\cos ADC
 
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\end{align*}
 
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</cmath>
 
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Since <math>\cos ADC = \cos (180 - ADB) = -\cos ADB</math>, we can add these two equations and get
 
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<cmath>5 = 10a^2</cmath>
 
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Hence <math>a = \frac{1}{\sqrt{2}}</math> and <math>BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}</math>.
 
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=== Solution 2 ===
 
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From [[Stewart's Theorem]], we have <math>(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.</math> Simplifying, we get <math>(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.</math>
 
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=== Solution 3 ===
 
  
 
<asy>
 
<asy>
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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>
 
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
+
~greenturtle 11/28/2017
 +
 
 +
=== Solution 2 ===
 +
 
 +
[[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
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 
 +
Since <math>\cos ADC = \cos (180 - ADB) = -\cos ADB</math>, we can add these two equations and get
 +
 
 +
<cmath>5 = 10a^2</cmath>
 +
 
 +
Hence <math>a = \frac{1}{\sqrt{2}}</math> and <math>BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}</math>.
 +
 
 +
=== Solution 3 ===
 +
 
 +
From [[Stewart's Theorem]], we have <math>(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.</math> Simplifying, we get <math>(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.</math>
  
 
== See also ==
 
== See also ==

Revision as of 23:38, 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

Solution 1

[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("$M$",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)); [/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 11/28/2017

Solution 2

2002 12B AMC-23.png

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 3

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

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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