Template:ProbAMC

Problem

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Solution

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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
}}