Difference between revisions of "1987 AHSME Problems/Problem 25"

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\textbf{(C)}\ \frac{3}{2} \qquad
 
\textbf{(C)}\ \frac{3}{2} \qquad
 
\textbf{(D)}\ \frac{13}{2}\qquad
 
\textbf{(D)}\ \frac{13}{2}\qquad
\textbf{(E)}\ \text{there is no minimum} </math>  
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\textbf{(E)}\ \text{there is no minimum} </math>  
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== Solution ==
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Let <math>C</math> have coordinates <math>(p, q)</math>. Then by the Shoelace Formula, the area of <math>\triangle ABC</math> is <math>\frac{3}{2} \lvert {12q-5p} \rvert</math>. Since <math>p</math> and <math>q</math> are integers, <math>\lvert {12q-5p} \rvert</math> is a positive integer, and by Bezout's Lemma, it can equal <math>1</math> (e.g. with <math>q = 2, p = 5</math>), so the minimum area is <math>\frac{3}{2} \times 1 = \frac{3}{2}</math>, which is answer <math>\boxed{C}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:56, 31 March 2018

Problem

$ABC$ is a triangle: $A=(0,0), B=(36,15)$ and both the coordinates of $C$ are integers. What is the minimum area $\triangle ABC$ can have?

$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \frac{13}{2}\qquad \textbf{(E)}\ \text{there is no minimum}$

Solution

Let $C$ have coordinates $(p, q)$. Then by the Shoelace Formula, the area of $\triangle ABC$ is $\frac{3}{2} \lvert {12q-5p} \rvert$. Since $p$ and $q$ are integers, $\lvert {12q-5p} \rvert$ is a positive integer, and by Bezout's Lemma, it can equal $1$ (e.g. with $q = 2, p = 5$), so the minimum area is $\frac{3}{2} \times 1 = \frac{3}{2}$, which is answer $\boxed{C}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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