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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>\overline{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 the area of the large triangle minus the area of the small triangle minus the area of the circle, which is <math>\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}</math>, which is approximately <math>51</math>. The answer is <math>\boxed{C}</math>.
+
<asy>
 +
Label f;
 +
f.p=fontsize(6);
 +
xaxis(-1,15,Ticks(f, 2.0));
 +
yaxis(-1,15,Ticks(f, 2.0));
 +
 
 +
pair A = MP("A",(2,2),SW), B = MP("B",(7,7),NE);
 +
D(A--B);
 +
filldraw((0,4)--(4,0)--(14,0)--(0,14)--cycle,gray);
 +
filldraw(CP(0.5(A+B),A),white);
 +
D(A);
 +
D(B);
 +
</asy>
 +
 
 +
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>\overline{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 the area of the large triangle minus the area of the small triangle minus the area of the circle, which is <math>\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}</math>, which is approximately <math>51</math>. The answer is <math>\boxed{\mathrm{C}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2005|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2005|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:20, 21 December 2020

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

[asy] Label f; f.p=fontsize(6); xaxis(-1,15,Ticks(f, 2.0)); yaxis(-1,15,Ticks(f, 2.0)); pair A = MP("A",(2,2),SW), B = MP("B",(7,7),NE); D(A--B); filldraw((0,4)--(4,0)--(14,0)--(0,14)--cycle,gray); filldraw(CP(0.5(A+B),A),white); D(A); D(B); [/asy]

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 $\overline{AB}$ 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 the area of the large triangle minus the area of the small triangle minus the area of the circle, which is $\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}$, which is approximately $51$. The answer is $\boxed{\mathrm{C}}$.

See also

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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
All AMC 12 Problems and Solutions

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