Difference between revisions of "1989 AIME Problems/Problem 15"
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=== Solution 4 === | === Solution 4 === | ||
− | First, let <math>[AEP]=a, [AFP]=b,</math> and <math>[ECP]=c.</math> Thus, we can easily find that <math>\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.</math> Now, <math>\frac{[ABP]}{BPD}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.</math> In the same manner, we find that <math>[CPD]=a+c.</math> Now, we can find that <math>\frac{[BPC]}{[PEC]}=\frac{9}{3}=3 \Leftrightarrow \frac{(3a)+(a+c)}{c}=3 \Leftrightarrow c=2a.</math> We can now use this to find that <math>\frac{[APC]}{[AFP]}=\frac{[BPC]}{[BFP]}=\frac{PC}{FP} \Leftrightarrow \frac{3a}{b}=\frac{6a}{3a-b} \Leftrightarrow a=b.</math> Plugging this value in, we find that <math>\frac{FC}{FP}=3 \Leftrightarrow PC=15, FP=5.</math> Now, since <math>\frac{[AEP]}{[PEC]}=\frac{a}{2a}=\frac{1}{2},</math> we can find that <math>2AE=EC.</math> Setting <math>AC=b,</math> we can apply Stewart's Theorem on triangle <math>APC</math> to find that <math>(15)(15)(\frac{b}{3})+(6)(6)(\frac{2b}{3})=(\frac{2b}{3})(\frac{b}{3})(b)+(b)(3)(3).</math> Solving, we find that <math>b=\sqrt{405} \Leftrightarrow AE=\frac{b}{3}=\sqrt{45}.</math> But, <math>3^2+6^2=45,</math> meaning that <math>\angle{APE}=90 \Leftrightarrow [APE]=\frac{(6)(3)}{2}=9=a.</math> Since <math>[ABC]=a+a+2a+2a+3a+3a=12a=(12)(9)=108,</math> we conclude that the answer is <math>\boxed{108}</math>. | + | First, let <math>[AEP]=a, [AFP]=b,</math> and <math>[ECP]=c.</math> Thus, we can easily find that <math>\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.</math> Now, <math>\frac{[ABP]}{BPD}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.</math> In the same manner, we find that <math>[CPD]=a+c.</math> Now, we can find that <math>\frac{[BPC]}{[PEC]}=\frac{9}{3}=3 \Leftrightarrow \frac{(3a)+(a+c)}{c}=3 \Leftrightarrow c=2a.</math> We can now use this to find that <math>\frac{[APC]}{[AFP]}=\frac{[BPC]}{[BFP]}=\frac{PC}{FP} \Leftrightarrow \frac{3a}{b}=\frac{6a}{3a-b} \Leftrightarrow a=b.</math> Plugging this value in, we find that <math>\frac{FC}{FP}=3 \Leftrightarrow PC=15, FP=5.</math> Now, since <math>\frac{[AEP]}{[PEC]}=\frac{a}{2a}=\frac{1}{2},</math> we can find that <math>2AE=EC.</math> Setting <math>AC=b,</math> we can apply Stewart's Theorem on triangle <math>APC</math> to find that <math>(15)(15)(\frac{b}{3})+(6)(6)(\frac{2b}{3})=(\frac{2b}{3})(\frac{b}{3})(b)+(b)(3)(3).</math> Solving, we find that <math>b=\sqrt{405} \Leftrightarrow AE=\frac{b}{3}=\sqrt{45}.</math> But, <math>3^2+6^2=45,</math> meaning that <math>\angle{APE}=90 \Leftrightarrow [APE]=\frac{(6)(3)}{2}=9=a.</math> Since <math>[ABC]=a+a+2a+2a+3a+3a=12a=(12)(9)=108,</math> we conclude that the answer is <math>\boxed{108.00}</math>. |
== See also == | == See also == |
Revision as of 12:26, 14 July 2021
Contents
Problem
Point is inside . Line segments , , and are drawn with on , on , and on (see the figure below). Given that , , , , and , find the area of .
Solution
Solution 1
Because we're given three concurrent cevians and their lengths, it seems very tempting to apply Mass points. We immediately see that , , and . Now, we recall that the masses on the three sides of the triangle must be balanced out, so and . Thus, and .
Recalling that , we see that and is a median to in . Applying Stewart's Theorem, , and . Now notice that , because both triangles share the same base and the . Applying Heron's formula on triangle with sides , , and , and .
Solution 2
Using a different form of Ceva's Theorem, we have
Solving and , we obtain and .
Let be the point on such that . Since and , . (Stewart's Theorem)
Also, since and , we see that , , etc. (Stewart's Theorem) Similarly, we have () and thus .
is a right triangle, so () is . Therefore, the area of . Using area ratio, .
Solution 3
Because the length of cevian is unknown, we can examine what happens when we extend it or decrease its length and see that it simply changes the angles between the cevians. Wouldn't it be great if it the length of was such that ? Let's first assume it's a right angle and hope that everything works out.
Extend to so that . The result is that , , and because . Now we see that if we are able to show that , that is , then our right angle assumption will be true.
Apply the Pythagorean Theorem on to get , so and . Now, we apply the Law of Cosines on triangles and .
Let . Notice that and , so we get two nice equations.
Solving, (yay!).
Now, the area is easy to find. .
[however, I think this solution is wrong, the A, B, and Cs do not match with the picture]
Solution 4
First, let and Thus, we can easily find that Now, In the same manner, we find that Now, we can find that We can now use this to find that Plugging this value in, we find that Now, since we can find that Setting we can apply Stewart's Theorem on triangle to find that Solving, we find that But, meaning that Since we conclude that the answer is .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Final Question | |
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.