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

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 $\boxed{\mathrm{C}}$ .

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

Latest revision as of 20:20, 21 December 2020

Problem

Solution

See also

@@ Line 1: / Line 1: @@
 == Problem ==
+Let <math>A(2,2)</math> and <math>B(7,7)</math> be points in the plane. Define <math>R</math> as the region in the first quadrant consisting of those points <math>C</math> such that <math>\triangle ABC</math> is an acute triangle. What is the closest integer to the area of the region <math>R</math>?
+<math>
+\mathrm{(A)}\ 25     \qquad
+\mathrm{(B)}\ 39     \qquad
+\mathrm{(C)}\ 51     \qquad
+\mathrm{(D)}\ 60      \qquad
+\mathrm{(E)}\ 80
+</math>
 == 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 <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 ==
-* [[2005 AMC 12B Problems]]
+{{AMC12 box|year=2005|ab=B|num-b=17|num-a=19}}
+{{MAA Notice}}