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2007 AMC 12B Problems/Problem 23: Difference between revisions

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And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>.
And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>.
\usepackage{tkz-euclide}
\usetkzobj{all}


==Solution #2==
==Solution #2==
Line 35: Line 38:
We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>.
We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>.


\begin{triangle}[thick]
\begin{tikzpicture}[thick]
\draw(0,0) -- (90:2cm) node[midway,left]{<math>opposite leg</math>} -- (0:4cm) node[midway,above right]{<math>hypotenuse</math>} -- (0,0) node[midway,below]{<math>adjacent leg</math>};
\coordinate (O) at (0,0);
\draw[fill=lightgray, thick] (0,0) -- (0:0.8cm) arc (0:90:0.8cm) node at (45:0.5cm) {<math>\gamma</math>} -- cycle;
\coordinate (A) at (4,0);
\end{triangle}
\coordinate (B) at (0,2);
\draw (O)--(A)--(B)--cycle;
 
\tkzLabelSegment[below=2pt](O,A){\textit{adjacent leg}}
\tkzLabelSegment[left=2pt](O,B){\textit{opposite leg}}
\tkzLabelSegment[above right=2pt](A,B){\textit{hypotenuse}}
 
\tkzMarkAngle[fill= orange,size=0.65cm,%
opacity=.4](A,O,B)
\tkzLabelAngle[pos = 0.35](A,O,B){<math>\gamma</math>}
 
\tkzMarkAngle[fill= orange,size=0.8cm,%
opacity=.4](B,A,O)
\tkzLabelAngle[pos = 0.6](B,A,O){<math>\alpha</math>}
 
\tkzMarkAngle[fill= orange,size=0.7cm,%
opacity=.4](O,B,A)
\tkzLabelAngle[pos = 0.5](O,B,A){<math>\beta</math>}
 
 
\end{tikzpicture}


==See Also==
==See Also==
{{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}}
{{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 12:46, 29 November 2014

Problem 23

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$

Solution

Let $a$ and $b$ be the two legs of the triangle.

We have $\frac{1}{2}ab = 3(a+b+c)$.

Then $ab=6\cdot (a+b+\sqrt {a^2 + b^2})$.

We can complete the square under the root, and we get, $ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})$.

Let $ab=p$ and $a+b=s$, we have $p=6\cdot (s+ \sqrt {s^2 - 2p})$.

After rearranging, squaring both sides, and simplifying, we have $p=12s-72$.


Putting back $a$ and $b$, and after factoring using $SFFT$, we've got $(a-12)\cdot (b-12)=72$.


Factoring 72, we get 6 pairs of $a$ and $b$


$(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).$


And this gives us $6$ solutions $\Rightarrow \mathrm{(A)}$.

\usepackage{tkz-euclide} \usetkzobj{all}

Solution #2

We will proceed by using the fact that $[ABC] = r\cdot s$, where $r$ is the radius of the incircle and $s$ is the semiperimeter ($s = \frac{p}{2}$).

We are given $[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6$.

\begin{tikzpicture}[thick] \coordinate (O) at (0,0); \coordinate (A) at (4,0); \coordinate (B) at (0,2); \draw (O)--(A)--(B)--cycle;

\tkzLabelSegment[below=2pt](O,A){\textit{adjacent leg}} \tkzLabelSegment[left=2pt](O,B){\textit{opposite leg}} \tkzLabelSegment[above right=2pt](A,B){\textit{hypotenuse}}

\tkzMarkAngle[fill= orange,size=0.65cm,% opacity=.4](A,O,B) \tkzLabelAngle[pos = 0.35](A,O,B){$\gamma$}

\tkzMarkAngle[fill= orange,size=0.8cm,% opacity=.4](B,A,O) \tkzLabelAngle[pos = 0.6](B,A,O){$\alpha$}

\tkzMarkAngle[fill= orange,size=0.7cm,% opacity=.4](O,B,A) \tkzLabelAngle[pos = 0.5](O,B,A){$\beta$}


\end{tikzpicture}

See Also

2007 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: Unable to save thumbnail to destination

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