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Difference between revisions of "2005 AMC 12B Problems/Problem 18"

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== Solution ==
 
== Solution ==
  
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For angle <math>A</math> and <math>B</math> to be acute, <math>C</math> must be between the two lines that are perpendicular to <math>\overline{AB}</math> and contain points <math>A</math> and <math>B</math>. For angle <math>C</math> to be acute, first draw a <math>45-45-90</math> triangle with <math>\bar{AB}</math> as the hypotenuse. Note <math>C</math> cannot be inside this triangle's circumscribed circle or else <math>\angle C > 90^\circ</math>. Hence, the area of <math>R</math> is
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For angle <math>A</math> and <math>B</math> to be acute, <math>C</math> must be between the two lines that are perpendicular to <math>\overline{AB}</math> and contain points <math>A</math> and <math>B</math>. For angle <math>C</math> to be acute, first draw a <math>45-45-90</math> triangle with <math>\bar{AB}</math> as the hypotenuse. Note <math>C</math> cannot be inside this triangle's circumscribed circle or else <math>\angle C > 90^\circ</math>. Hence, the area of <math>R</math> is <math>\frac{2}{2}</math>.
  
 
== See also ==
 
== See also ==
 
* [[2005 AMC 12B Problems]]
 
* [[2005 AMC 12B Problems]]

Revision as of 18:52, 12 September 2010

Problem

Let $A(2,2)$ and $B(7,7)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R$?

$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 39 \qquad \mathrm{(C)}\ 51 \qquad \mathrm{(D)}\ 60 \qquad \mathrm{(E)}\ 80$

Solution

For angle $A$ and $B$ to be acute, $C$ must be between the two lines that are perpendicular to $\overline{AB}$ and contain points $A$ and $B$. For angle $C$ to be acute, first draw a $45-45-90$ triangle with $\bar{AB}$ as the hypotenuse. Note $C$ cannot be inside this triangle's circumscribed circle or else $\angle C > 90^\circ$. Hence, the area of $R$ is $\frac{2}{2}$.

See also

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