# Template:ProbAMC

## Contents

## Problem

*This problem has not been edited in. If you know this problem, please help us out by adding it.*

## Solution

*This problem needs a solution. If you have a solution for it, please help us out by adding it.*

## See Also

{{AMC{{{tentwelve}}} box|year={{{year}}}|ab=|num-b={{{num-b}}}|num-a={{{num-a}}}|before|after}}

[[Category:{{{diff}}} {{{subject}}} Problems]]

# Documentation

Use it as follows:

{{subst:probAMC | problem = | answera = | answerb = | answerc = | answerd = | answere = | solution = | tentwelve = | year = | ab = | num-b = | num-a = | diff = | subject = | before = | after = }}

The parameters should be filled as follows:

{{subst:probAMC | problem = The problem statement goes here. | answera = The first answer choice | answerb = The second | answerc = .. | answerd = .. | answere = The fifth | solution = The solution goes here. | tentwelve = 10 or 12 | year = YYYY | ab = A or B, if applicable | num-b = The number before this problem, ex if this is #5, write 4 | num-a = The number after this problem, ex if this is #5, write 6 | diff = Introductory or Intermediate | subject = Algebra or Number Theory or Geometry or Combinatorics or Trigonometry | before = before = First problem <!-- If this is the first problem, fill in this, else leave blank --> | after = after = Last problem <!-- If this is the last problem, fill in this, else leave blank --> }}

As an example, (see 2002 AMC 12B Problems/Problem 24)

{{subst:probAMC | problem = A [[convex polygon|convex]] [[quadrilateral]] $ABCD$ with area $2002$ contains a point $P$ in its interior such that $PA = 24, PB = 32, PC = 28, PD = 45$. Find the perimeter of $ABCD$. | answera = 4\sqrt{2002} | answerb = 2\sqrt{8465} | answerc = 2(48+\sqrt{2002}) | answerd = 2\sqrt{8633} | answere = 4(36 + \sqrt{113}) | solution =We have $$[ABCD] = 2002 \le \frac 12 (AC \cdot BD)$$ (Why is this true? Try splitting the quadrilateral along $AC$ and then using the triangle area formula), with equality if $\overline{AC} \perp \overline{BD}$. By the [[triangle inequality]], $$\begin{align*}AC &\le PA + PC = 52\\ PD &\le PB + PD = 77\end{align*}$$ with equality if $P$ lies on $\overline{AC}$ and $\overline{BD}$ respectively. Thus $$2002 \le \frac{1}{2} AC \cdot BD \le \frac 12 \cdot 52 \cdot 77 = 2002$$ Therefore $\overline{AC} \perp \overline{BD}$ at point $P$. [[Image:2002_12B_AMC-24.png|center]] By the [[Pythagorean Theorem]], $$\begin{align*} AB = \sqrt{PA^2 + PB^2} = \sqrt{24^2 + 32^2} = 40\\ BC = \sqrt{PB^2 + PC^2} = \sqrt{32^2 + 28^2} = 4\sqrt{113}\\ CD = \sqrt{PC^2 + PD^2} = \sqrt{28^2 + 45^2} = 53\\ DA = \sqrt{PD^2 + PA^2} = \sqrt{45^2 + 24^2} = 51 \end{align*}$$ The perimeter of $ABCD$ is $AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}$. | tentwelve = 12 | year = 2002 | ab = B | num-b = 23 | num-a = 25 | diff = Introductory | subject = Geometry }}