Difference between revisions of "2003 AMC 12A Problems/Problem 16"
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===Solution 2=== | ===Solution 2=== | ||
− | We will use geometric probability. Let us take point P, and draw the perpendiculars to BC, CA, and AB, and call the feet of these perpendiculars D, E, and F respectively. The area of triangle ACP is simply <math>\frac{1}{2} * AC * PF</math>. Similarly we can find the area of triangles BCP and ABP. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose P, PD + PE + PF = the height of the triangle. Setting the area of triangle ABP greater than ACP and BCP, we want PF to be the largest of PF, PD, and PE. We then realize that PF = PD = PE when P is the incenter of ABC. Let us call the incenter of the triangle Q. If we want PF to be the largest of the three, by testing points we realize that P must be in the interior of quadrilateral QDCE. So our probability (using geometric probability) is the area of QDCE divided by the area of ABC. We will now show that the three quadrilaterals, QDCE, QEAF, and QFBD are congruent. As the definition of point Q yields, QF = QD = QE. Since ABC is equilateral, Q is also the circumcenter of ABC, so QA = QB = QC. By the Pythagorean Theorem, BD = DC = CE = EA = AF = FB. Also, angles BDQ, BFQ, CEQ, CDQ, AFQ, and AEQ are all equal to 90 | + | 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>. |
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+ | ==Solution 3 (Solution 1 but for dumb dumbs)== | ||
+ | Due to symmetry, we notice that the probability one triangle is greater than the other is <math>\frac{1}{2}</math>. For two triangles this gives <math>\frac{1}{2} * \frac{1}{2} = \frac{1}{4}</math>. However, the probability of one triangle being greater than the other, depends on knowing whether one triangle is greater or less than the other. This means the probability is less than <math>\frac{1}{2}</math> but greater than <math>\frac{1}{4}</math> giving <math>\boxed{\mathrm{(C)}\ \dfrac{1}{3}}</math>. | ||
+ | |||
+ | ~PeterDoesPhysics | ||
==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 19:57, 27 August 2024
Contents
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 .
Solution 3 (Solution 1 but for dumb dumbs)
Due to symmetry, we notice that the probability one triangle is greater than the other is . For two triangles this gives . However, the probability of one triangle being greater than the other, depends on knowing whether one triangle is greater or less than the other. This means the probability is less than but greater than giving .
~PeterDoesPhysics
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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