# Template:ProbAMC

## Problem $\mathrm{(A)}\ {{{answera}}} \qquad \mathrm{(B)}\ {{{answerb}}} \qquad \mathrm{(C)}\ {{{answerc}}} \qquad \mathrm{(D)}\ {{{answerd}}} \qquad \mathrm{(E)}\ {{{answere}}}$

## Solution

{{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 =
| 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.
| 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 $ABCD$.
| 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
}}