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

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\qquad\mathrm{(E)}\ \sqrt{3}</math>
 
\qquad\mathrm{(E)}\ \sqrt{3}</math>
 
== Solution ==
 
== Solution ==
 +
 +
=== Solution 1: Pythagorean Theorem ===
 +
 +
<asy>
 +
 +
unitsize(4cm);
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 +
pair A, B, C, D, M;
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 +
A = (1.768,0.935);
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B = (1.414,0);
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C = (0,0);
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D = (1.768,0);
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M = (0.707,0);
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draw(A--B--C--cycle);
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draw(A--D);
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draw(D--B);
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draw(A--M);
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label("$A$",A,N);
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label("$B$",B,S);
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label("$C$",C,S);
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label("$D$",D,S);
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label("$M$",M,S);
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label("$x$",(A+D)/2,E);
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label("$y$",(B+D)/2,S);
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label("$a$",(C+M)/2,S);
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label("$a$",(M+B)/2,S);
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label("$2a$",(A+M)/2,SE);
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label("$1$",(A+B)/2,SE);
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label("$2$",(A+C)/2,NW);
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draw(rightanglemark(B,D,A,3));
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</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>.
 +
Using the Pythagorean Theorem, we obtain the equations
 +
 +
<cmath>
 +
\begin{align*}
 +
x^2 + y^2 = 1 \hspace{0.5cm}(1)\\
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x^2 + y^2 + 2ya + a^2 = 4a^2 \hspace{0.5cm}(2)\\
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x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3)
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\end{align*}
 +
</cmath>
 +
 +
Subtracting <math>(1)</math> equation from <math>(2)</math> and <math>(3)</math>, we get
 +
 +
<cmath>
 +
\begin{align*}
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2ya + a^2 = 4a^2 - 1 \hspace{0.5cm}(4)\\
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4ya + 4a^2 = 3 \hspace{0.5cm}(5)
 +
\end{align*}
 +
</cmath>
 +
 +
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 11/28/2017
 +
 +
=== Solution 2: Law of Cosines ===
  
 
[[Image:2002_12B_AMC-23.png]]
 
[[Image:2002_12B_AMC-23.png]]
−
 
−
=== Solution 1 ===
 
  
 
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  
 
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  
Line 27: Line 87:
 
Hence <math>a = \frac{1}{\sqrt{2}}</math> and <math>BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}</math>.
 
Hence <math>a = \frac{1}{\sqrt{2}}</math> and <math>BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}</math>.
  
−
=== Solution 2 ===
+
=== Solution 3: Stewart's Theorem ===
  
 
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>
 
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>
 +
- awu2014
  
−
=== Solution 3 ===
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=== Solution 4: Pappus's Median Theorem ===
  
−
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>.
+
There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have <math>\triangle{ABC}</math>, and you draw a median from point <math>A</math> to side <math>BC</math> (label this as <math>M</math>), then: <math>(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}</math>. Note that <math>b</math> is the length of side <math>\overline{AC}</math>, <math>c</math> is the length of side <math>\overline{AB}</math>, and <math>a</math> is length of side <math>\overline{BC}</math>. Let <math>MB = MC = x</math>. Then <math>AM = 2x</math>. Now, we can plug into the formula given above: <math>AM = 2x</math>, <math>b = 2</math>, <math>c = 1</math>, and <math>a = 2x</math>. After some simple algebra, we find <math>x = \dfrac{\sqrt{2}}{2}</math>. Then, <math>BC = \boxed{\sqrt{2}} \implies \boxed{C}</math>.
−
Using the Pythagorean Theorem, we obtain the equations
 
  
−
<cmath>
+
-Flames
−
\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*}
 
−
</cmath>
 
  
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Subtracting the first equation from the second and third equations, we get
+
Note: Pappus's Median Theorem is just a special case of Stewart's Theorem, with <math>m = n</math>. ~Puck_0
 +
aka Apollonius' Theorem - Orion 2010
  
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<cmath>
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===Video Solution by TheBeautyofMath===
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\begin{align*}
+
https://youtu.be/jEVMgWKQIW8
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2ya + a^2 = 4a^2 - 1 \\
 
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4ya + 4a^2 = 3
 
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\end{align*}
 
−
</cmath>
 
  
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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}\Rightarrow \boxed{\mathrm{(C)}}</math>
+
~IceMatrix
  
 
== See also ==
 
== See also ==

Latest revision as of 12:53, 9 July 2024

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: Pythagorean Theorem

[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: Law of Cosines

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: Stewart's Theorem

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 $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.$ - awu2014

Solution 4: Pappus's Median Theorem

There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have $\triangle{ABC}$, and you draw a median from point $A$ to side $BC$ (label this as $M$), then: $(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}$. Note that $b$ is the length of side $\overline{AC}$, $c$ is the length of side $\overline{AB}$, and $a$ is length of side $\overline{BC}$. Let $MB = MC = x$. Then $AM = 2x$. Now, we can plug into the formula given above: $AM = 2x$, $b = 2$, $c = 1$, and $a = 2x$. After some simple algebra, we find $x = \dfrac{\sqrt{2}}{2}$. Then, $BC = \boxed{\sqrt{2}} \implies \boxed{C}$.

-Flames

Note: Pappus's Median Theorem is just a special case of Stewart's Theorem, with $m = n$. ~Puck_0 aka Apollonius' Theorem - Orion 2010

Video Solution by TheBeautyofMath

https://youtu.be/jEVMgWKQIW8

~IceMatrix

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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