Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2002 AMC 12B Problems/Problem 23: Difference between revisions

← Older editNewer edit →
Greenturtle (talk | contribs)
editor
25 edits
Greenturtle (talk | contribs)
editor
25 edits
Line 38: Line 38:
<cmath>
<cmath>
\begin{align*}
\begin{align*}
  x^2 + y^2 = 1 \\
  x^2 + y^2 = 1 (1)\\
x^2 + y^2 + 2ya + a^2 = 4a^2 \\
x^2 + y^2 + 2ya + a^2 = 4a^2 (2)\\
x^2 + y^2 + 4ya + 4a^2 = 4
x^2 + y^2 + 4ya + 4a^2 = 4 (3)
\end{align*}
\end{align*}
</cmath>
</cmath>


Subtracting the first equation from the second and third equations, we get
Subtracting (1) equation from (2) and (3), we get


<cmath>
<cmath>
\begin{align*}
\begin{align*}
2ya + a^2 = 4a^2 - 1 \\
2ya + a^2 = 4a^2 - 1 (4)\\
4ya + 4a^2 = 3
4ya + 4a^2 = 3 (5)
\end{align*}
\end{align*}
</cmath>
</cmath>


Then, subtracting two times the first equation from the second and rearranging, we get <math>10a^2 = 5</math>, so <math>BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}</math>
Then, subtracting two times (4) from (5) 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 11/26/2017

Revision as of 23:30, 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

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 (1)\\ x^2 + y^2 + 2ya + a^2 = 4a^2 (2)\\ x^2 + y^2 + 4ya + 4a^2 = 4 (3) \end{align*}

Subtracting (1) equation from (2) and (3), we get

\begin{align*} 2ya + a^2 = 4a^2 - 1 (4)\\ 4ya + 4a^2 = 3 (5) \end{align*}

Then, subtracting two 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_23&oldid=88671"