Difference between revisions of "2003 AMC 12A Problems/Problem 16"
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== Problem == | == Problem == | ||
− | A point P is chosen at random in the interior of equilateral triangle <math>ABC</math>. What is the probability that <math>\triangle ABP</math> has a greater area than each of <math>\triangle ACP</math> and <math>\triangle BCP</math>? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>A point P is chosen at random in the interior of equilateral triangle <math>ABC</math>. What is the probability that <math>\triangle ABP</math> has a greater area than each of <math>\triangle ACP</math> and <math>\triangle BCP</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{3} </math> | <math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{3} </math> | ||
== Solution== | == Solution== | ||
+ | <asy> | ||
+ | draw((0,10)--(8.660254037844385792,-5)--(-8.660254037844385792,-5)--cycle); | ||
+ | dot((0,0)); | ||
+ | label("$P$",(0,0),N); | ||
+ | </asy> | ||
===Solution 1=== | ===Solution 1=== | ||
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===Solution 2=== | ===Solution 2=== | ||
− | We will use | + | We will use geometric probability. Let us take point <math>P</math>, and draw the perpendiculars to <math>BC</math>, <math>CA</math>, and <math>AB</math>, and call the feet of these perpendiculars <math>D</math>, <math>E</math>, and <math>F</math> respectively. The area of <math>\triangle ACP</math> is simply <math>\frac{1}{2} * AC * PF</math>. Similarly we can find the area of triangles <math>BCP</math> and <math>ABP</math>. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose <math>P, PD + PE + PF</math> = the height of the triangle. Setting the area of triangle <math>ABP</math> greater than <math>ACP</math> and <math>BCP</math>, we want <math>PF</math> to be the largest of <math>PF</math>, <math>PD</math>, and <math>PE</math>. We then realize that <math>PF = PD = PE</math> when <math>P</math> is the incenter of <math>\triangle ABC</math>. Let us call the incenter of the triangle <math>Q</math>. If we want <math>PF</math> to be the largest of the three, by testing points we realize that <math>P</math> must be in the interior of quadrilateral <math>QDCE</math>. So our probability (using geometric probability) is the area of <math>QDCE</math> divided by the area of <math>ABC</math>. We will now show that the three quadrilaterals, <math>QDCE</math>, <math>QEAF</math>, and <math>QFBD</math> are congruent. As the definition of point <math>Q</math> yields, <math>QF</math> = <math>QD</math> = <math>QE</math>. Since <math>ABC</math> is equilateral, <math>Q</math> is also the circumcenter of <math>\triangle ABC</math>, so <math>QA = QB = QC</math>. By the Pythagorean Theorem, <math>BD = DC = CE = EA = AF = FB</math>. Also, angles <math>BDQ, BFQ, CEQ, CDQ, AFQ</math>, and <math>AEQ</math> are all equal to <math>90^\circ</math>. Angles <math>DBF, FAE, ECD</math> are all equal to <math>60</math> degrees, so it is now clear that quadrilaterals <math>QDCE, QEAF, QFBD</math> are all congruent. Summing up these areas gives us the area of <math>\triangle ABC</math>. <math>QDCE</math> contributes to a third of that area so <math>\frac{[QDCE]}{[ABC]}=\boxed{\mathrm{(C)}\ \dfrac{1}{3}}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2003|ab=A|num-b=15|num-a=17}} | {{AMC12 box|year=2003|ab=A|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:27, 6 July 2017
Problem
A point P is chosen at random in the interior of equilateral triangle . What is the probability that has a greater area than each of and ?
Solution
Solution 1
After we pick point , we realize that is symmetric for this purpose, and so the probability that is the greatest area, or or , are all the same. Since they add to , the probability that has the greatest area is
Solution 2
We will use geometric probability. Let us take point , and draw the perpendiculars to , , and , and call the feet of these perpendiculars , , and respectively. The area of is simply . Similarly we can find the area of triangles and . If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose = the height of the triangle. Setting the area of triangle greater than and , we want to be the largest of , , and . We then realize that when is the incenter of . Let us call the incenter of the triangle . If we want to be the largest of the three, by testing points we realize that must be in the interior of quadrilateral . So our probability (using geometric probability) is the area of divided by the area of . We will now show that the three quadrilaterals, , , and are congruent. As the definition of point yields, = = . Since is equilateral, is also the circumcenter of , so . By the Pythagorean Theorem, . Also, angles , and are all equal to . Angles are all equal to degrees, so it is now clear that quadrilaterals are all congruent. Summing up these areas gives us the area of . contributes to a third of that area so .
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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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