1989 AIME Problems/Problem 15
Contents
- 1 Problem
- 2 Solutions
- 2.1 Solution 1 (Ceva's Theorem, Stewart's Theorem)
- 2.2 Solution 2 (Mass Points, Stewart's Theorem, Heron's Formula)
- 2.3 Solution 3 (Ceva's Theorem, Stewart's Theorem)
- 2.4 Solution 4 (Stewart's Theorem)
- 2.5 Solution 5 (Mass of a Point, Stewart's Theorem, Heron's Formula)
- 2.6 Solution 6 (easier version of Solution 5)
- 2.7 Solution 7 (Mass Points, Stewart's Theorem, Simple Version)
- 2.8 Solution 8 (Ratios, Auxiliary Lines and 3-4-5 triangle)
- 2.9 Solution 9 (Just Trig Bash)
- 3 See also
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 .
Solutions
Solution 1 (Ceva's Theorem, Stewart's Theorem)
Let be the area of polygon . We'll make use of the following fact: if is a point in the interior of triangle , and line intersects line at point , then
This is true because triangles and have their areas in ratio (as they share a common height from ), and the same is true of triangles and .
We'll also use the related fact that . This is slightly more well known, as it is used in the standard proof of Ceva's theorem.
Now we'll apply these results to the problem at hand.
Since , this means that ; thus has half the area of . And since , we can conclude that has one third of the combined areas of triangle and , and thus of the area of . This means that is left with of the area of triangle : Since , and since , this means that is the midpoint of .
Furthermore, we know that , so .
We now apply Stewart's theorem to segment in —or rather, the simplified version for a median. This tells us that Plugging in we know, we learn that Happily, is also equal to 117. Therefore is a right triangle with a right angle at ; its area is thus . As is a median of , the area of is twice this, or 54. And we already know that has half the area of , which must therefore be .
Solution 2 (Mass Points, Stewart's Theorem, Heron's Formula)
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 3 (Ceva's Theorem, Stewart's Theorem)
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 4 (Stewart's Theorem)
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 .
Solution 5 (Mass of a Point, Stewart's Theorem, Heron's Formula)
Firstly, since they all meet at one single point, denoting the mass of them separately. Assuming ; we can get that ; which leads to the ratio between segments, Denoting that
Now we know three cevians' length, Applying Stewart theorem to them, getting three different equations: After solving the system of equation, we get that ;
pulling back to get the length of ; now we can apply Heron's formula here, which is
Our answer is .
~bluesoul
Note (how to find x and y without the system of equations)
To ease computation, we can apply Stewart's Theorem to find , , and directly. Since and , and . We can apply Stewart's Theorem on to get . Solving, we find that . We can do the same on and to obtain and . We proceed with Heron's Formula as the solution states.
~kn07
Solution 6 (easier version of Solution 5)
In Solution 5, instead of finding all of , we only need . This is because after we solve for , we can notice that is isosceles with . Because is the midpoint of the base, is an altitude of . Therefore, . Using the same altitude property, we can find that .
-NL008
Solution 7 (Mass Points, Stewart's Theorem, Simple Version)
Set and use mass points to find that and Using Stewart's Theorem on we find that Then we notice that is right, which means the area of is Because the area of is times the area of which means the area of
Solution 8 (Ratios, Auxiliary Lines and 3-4-5 triangle)
We try to solve this using only elementary concepts. Let the areas of triangles , and be , and respectively. Then and . Hence . Similarly and since we then have . Additionally we now see that triangles and are similar, so and . Hence . Now construct a point on segment such that and , we will have , and hence , giving . Triangle is therefore a 3-4-5 triangle! So and so . Then it is easy to calculate that and the area of triangle . ~Leole
Solution 9 (Just Trig Bash)
We start with mass points as in Solution 2, and receive , , . Law of Cosines on triangles and with and gives Adding them: , so . Similarly, and . Using Heron's,
~sml1809
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 |
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