Difference between revisions of "2013 AIME I Problems/Problem 7"
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==Solution Pythagorean== | ==Solution Pythagorean== | ||
− | Let half the height be <math>a</math> (we want to find <math>2a</math>), then we see that the three sides of the triangle are (by Pyth Theorem) <math>10, \sqrt{a^2+36}, \sqrt{a^2+64}</math>. Using the Law of Sines with the angle as the one included between the square roots, we see that this angle's cosine is <math>\frac{a^2}{\sqrt{(a^2+36)(a^2+64)}</math> by the Law of Cosines, meaning that its sine is <math>\frac{\sqrt{100a^2+2304}}{\sqrt{(a^2+36)(a^2+64)}}. | + | Let half the height be <math>a</math> (we want to find <math>2a</math>), then we see that the three sides of the triangle are (by Pyth Theorem) <math>10, \sqrt{a^2+36}, \sqrt{a^2+64}</math>. Using the Law of Sines with the angle as the one included between the square roots, we see that this angle's cosine is <math>\frac{a^2}{\sqrt{(a^2+36)(a^2+64)}}</math> by the Law of Cosines, meaning that its sine is <math>\frac{\sqrt{100a^2+2304}}{\sqrt{(a^2+36)(a^2+64)}}. |
Now, note that </math>\sqrt{25a^2+576} = 30<math>, giving </math>a=\frac{18}{5}<math>, so our answer is </math>\boxed{041}$. | Now, note that </math>\sqrt{25a^2+576} = 30<math>, giving </math>a=\frac{18}{5}<math>, so our answer is </math>\boxed{041}$. |
Revision as of 20:42, 23 January 2018
Contents
Problem 7
A rectangular box has width inches, length inches, and height inches, where and are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of square inches. Find .
Solution 1
Let the height of the box be .
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, , and . Since the area of the triangle is , the altitude of the triangle from the base with length is .
Considering the two triangles created by the altitude, we use the Pythagorean theorem twice to find the lengths of the two line segments that make up the base of .
We find:
Solving for gives us . Since this fraction is simplified:
Solution 2
We may use vectors. Let the height of the box be . Without loss of generality, let the front bottom left corner of the box be . Let the center point of the bottom face be , the center of the left face be and the center of the front face be .
We are given that the area of the triangle is . Thus, by a well known formula, we note that We quickly attain that and (We can arbitrarily assign the long and short ends due to symmetry)
Computing the cross product, we find:
Thus:
Solution 3
Let the height of the box be .
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, , and . Therefore, we can use Heron's formula to set up an equation for the area of the triangle.
The semiperimeter is . Therefore, when we square Heron's formula, we find
Solving, we get .
Solution 4
It isn't hard to see that the triangle connecting the centers of the faces of the rectangular prism is congruent to the triangle connecting the midpoints of three edges that concur. So we can now apply de Guas theorem to see that:
Where is half the desired length of the height.
Solving yields
And thus
---Solution 4 contributed by Siddharth Namachivayam
Solution Pythagorean
Let half the height be (we want to find ), then we see that the three sides of the triangle are (by Pyth Theorem) . Using the Law of Sines with the angle as the one included between the square roots, we see that this angle's cosine is by the Law of Cosines, meaning that its sine is $\frac{\sqrt{100a^2+2304}}{\sqrt{(a^2+36)(a^2+64)}}.
Now, note that$ (Error compiling LaTeX. ! Missing $ inserted.)\sqrt{25a^2+576} = 30a=\frac{18}{5}\boxed{041}$.
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.