Difference between revisions of "2017 AMC 10A Problems/Problem 17"

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<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 3\sqrt{5}\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 5\sqrt{2}</math>
 
<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 3\sqrt{5}\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 5\sqrt{2}</math>
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==Solution==
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Because <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are integers there are only a few coordinates that actually satisfy the equation. The coordinates are <math>(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),</math> and <math>(\pm 5,0).</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:14, 8 February 2017

Problem

Distinct points $P$, $Q$, $R$, $S$ lie on the circle $x^2+y^2=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$?

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 3\sqrt{5}\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 5\sqrt{2}$

Solution

Because $P$, $Q$, $R$, and $S$ are integers there are only a few coordinates that actually satisfy the equation. The coordinates are $(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),$ and $(\pm 5,0).$

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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