Difference between revisions of "1987 AHSME Problems/Problem 25"
(Created page with "==Problem== <math>ABC</math> is a triangle: <math>A=(0,0), B=(36,15)</math> and both the coordinates of <math>C</math> are integers. What is the minimum area <math>\triangle ABC...") |
(Added a solution with explanation) |
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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> | + | \textbf{(E)}\ \text{there is no minimum} </math> |
+ | |||
+ | == Solution == | ||
+ | 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 11:56, 31 March 2018
Problem
is a triangle: and both the coordinates of are integers. What is the minimum area can have?
Solution
Let have coordinates . Then by the Shoelace Formula, the area of is . Since and are integers, is a positive integer, and by Bezout's Lemma, it can equal (e.g. with ), so the minimum area is , which is answer .
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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