Difference between revisions of "1988 AIME Problems/Problem 12"
I like pie (talk | contribs) (→Solution) |
|||
(5 intermediate revisions by 4 users not shown) | |||
Line 4: | Line 4: | ||
[[Image:1988_AIME-12.png]] | [[Image:1988_AIME-12.png]] | ||
− | == Solution == | + | == Solution 1 == |
Call the [[cevian]]s AD, BE, and CF. Using area ratios (<math>\triangle PBC</math> and <math>\triangle ABC</math> have the same base), we have: | Call the [[cevian]]s AD, BE, and CF. Using area ratios (<math>\triangle PBC</math> and <math>\triangle ABC</math> have the same base), we have: | ||
Line 22: | Line 22: | ||
<cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath> | <cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath> | ||
<cmath>9(a + b + c) + 54 = abc=\boxed{441}</cmath> | <cmath>9(a + b + c) + 54 = abc=\boxed{441}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>A,B,C</math> be the weights of the respective vertices. We see that the weights of the feet of the cevians are <math>A+B,B+C,C+A</math>. By [[mass points]], we have that: <cmath>\dfrac{a}{3}=\dfrac{B+C}{A}</cmath> <cmath>\dfrac{b}{3}=\dfrac{C+A}{B}</cmath> <cmath>\dfrac{c}{3}=\dfrac{A+B}{C}</cmath> | ||
+ | |||
+ | If we add the equations together, we get <math>\frac{a+b+c}{3}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\frac{43}{3}</math> | ||
+ | |||
+ | If we multiply them together, we get <math>\frac{abc}{27}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B+2ABC}{ABC}=\frac{49}{3} \implies abc=\boxed{441}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | You can use mass points to derive <math>\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d}=1.</math> Plugging it in yields <math>\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1.</math> We proceed as we did in Solution 1 - however, to make the equation look less messy, we do the substitution <math>a'=a+3,b'=b+3,c'=c+3.</math> | ||
+ | |||
+ | Then we have <math>\frac{3}{a'}+\frac{3}{b'}+\frac{3}{c'}=1.</math> Clearing fractions gives us <math>a'b'c'=3a'b'+3b'c'+3c'a'\to a'b'c'-3a'b'-3b'c'-3c'a'=0.</math> Factoring yields <math>(a'-3)(b'-3)(c'-3)=9(a'+b'+c')-27,</math> and the left hand side looks suspiciously like what we want to find. (It is.) | ||
+ | |||
+ | Substituting yields our answer as <math>9\cdot 52-27=\boxed{441}.</math> | ||
== See also == | == See also == | ||
Line 27: | Line 43: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:41, 30 December 2018
Problem
Let be an interior point of triangle and extend lines from the vertices through to the opposite sides. Let , , , and denote the lengths of the segments indicated in the figure. Find the product if and .
Solution 1
Call the cevians AD, BE, and CF. Using area ratios ( and have the same base), we have:
Similarily, and .
Then,
The identity is a form of Ceva's Theorem.
Plugging in , we get
Solution 2
Let be the weights of the respective vertices. We see that the weights of the feet of the cevians are . By mass points, we have that:
If we add the equations together, we get
If we multiply them together, we get
Solution 3
You can use mass points to derive Plugging it in yields We proceed as we did in Solution 1 - however, to make the equation look less messy, we do the substitution
Then we have Clearing fractions gives us Factoring yields and the left hand side looks suspiciously like what we want to find. (It is.)
Substituting yields our answer as
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.