Difference between revisions of "2014 AIME I Problems/Problem 13"
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== Solution 3== | == Solution 3== | ||
− | Continue in the same way as solution 1 to get that POK has area 3a, and OK = d/ | + | Continue in the same way as solution 1 to get that <math>POK</math> has area <math>3a</math>, and <math>OK = \frac{d}{10}</math>. You can then find <math>PK</math> has length <math>\frac 32</math>. |
− | Then, if we drop a perpendicular from H to BC | + | Then, if we drop a perpendicular from <math>H</math> to <math>BC</math> at <math>L</math>, We get <math>\triangle HLF \sim \triangle OPK</math>. |
− | Thus, <math>LF = \frac{15\cdot 34}{d}</math>, and we know HL = d, and HF = 34. Thus, we can set up an equation in terms of d using the Pythagorean theorem. | + | Thus, <math>LF = \frac{15\cdot 34}{d}</math>, and we know <math>HL = d</math>, and <math>HF = 34</math>. Thus, we can set up an equation in terms of <math>d</math> using the Pythagorean theorem. |
<cmath>\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2</cmath> | <cmath>\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2</cmath> |
Revision as of 02:53, 25 February 2021
Problem 13
On square , points , and lie on sides and respectively, so that and . Segments and intersect at a point , and the areas of the quadrilaterals and are in the ratio Find the area of square .
Solution
Notice that . This means passes through the center of the square.
Draw with on , on such that and intersects at the center of the square which I'll label as .
Let the area of the square be . Then the area of and the area of . This is because is perpendicular to (given in the problem), so is also perpendicular to . These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.
Let the side length of the square be .
Draw and intersects at . .
The area of , so the area of .
Let . Then
Consider the area of .
Thus, .
Solving , we get .
Therefore, the area of
Lazy Solution
, a multiple of . In addition, , which is . Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to and must be a multiple of . All of these triples are primitive:
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting :
Thus, is the only valid answer.
Solution 3
Continue in the same way as solution 1 to get that has area , and . You can then find has length .
Then, if we drop a perpendicular from to at , We get .
Thus, , and we know , and . Thus, we can set up an equation in terms of using the Pythagorean theorem.
is extraneous, so . Since the area is , we have it is equal to
-Alexlikemath
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.