Difference between revisions of "1991 AIME Problems/Problem 12"

 
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== Problem ==
 
== Problem ==
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Rhombus <math>PQRS^{}_{}</math> is inscribed in rectangle <math>ABCD^{}_{}</math> so that vertices <math>P^{}_{}</math>, <math>Q^{}_{}</math>, <math>R^{}_{}</math>, and <math>S^{}_{}</math> are interior points on sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, and <math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{}_{}=40</math>. Let <math>m/n^{}_{}</math>, in lowest terms, denote the perimeter of <math>ABCD^{}_{}</math>. Find <math>m+n^{}_{}</math>.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
  
 
== See also ==
 
== See also ==
* [[1991 AIME Problems]]
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{{AIME box|year=1991|num-b=11|num-a=13}}

Revision as of 02:38, 2 March 2007

Problem

Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$, $Q^{}_{}$, $R^{}_{}$, and $S^{}_{}$ are interior points on sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. It is given that $PB^{}_{}=15$, $BQ^{}_{}=20$, $PR^{}_{}=30$, and $QS^{}_{}=40$. Let $m/n^{}_{}$, in lowest terms, denote the perimeter of $ABCD^{}_{}$. Find $m+n^{}_{}$.

Solution

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See also

1991 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions