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

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

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

@@ Line 9: / Line 9: @@
 == Solution ==
-=== Solution 1 ===
+=== Solution 1: Pythagorean Theorem ===
 <asy>
@@ Line 69: / Line 69: @@
 ~greenturtle 11/28/2017
-=== Solution 2 ===
+=== Solution 2: Law of Cosines ===
 [[Image:2002_12B_AMC-23.png]]
@@ Line 87: / Line 87: @@
 Hence <math>a = \frac{1}{\sqrt{2}}</math> and <math>BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}</math>.
-=== Solution 3 ===
+=== 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>
 - awu2014
-=== Solution 4 [Pappus's Median Theorem] ===
+=== Solution 4: Pappus's Median Theorem ===
-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 = \sqrt{2}</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>.
 -Flames
+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
 ===Video Solution by TheBeautyofMath===

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