2018 AMC 10A Problems/Problem 23
Problem
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from to the hypotenuse is 2 units. What fraction of the field is planted?
Solution
Let the square have side length . Connect the upper-right vertex of square with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is .
Square has area , and the two thin triangle regions have area and . The final triangular region with the hypotenuse as its base and height has area . Thus, we have
Solving gives . The area of is and the desired ratio is .
Solution 2
Let the square have side length . If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is . Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is , and using the ratios of the side lengths, the height is . Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so
Now comes the easy part: finding the ratio of the areas:
Solution 3
We use coordinate geometry. Let the right angle be at and the hypotenuse be the line for . Denote the position of as , and by the point to line distance formula, we know that Obviously , so , and from here the rest of the solution follows to get .
Solution 4
Let the side length of the square be . First off, let us make a similar triangle with the segment of length and the top-right corner of . Therefore, the longest side of the smaller triangle must be . We then do operations with that side in terms of . We subtract from the bottom, and from the top. That gives us the equation of . Solving,
Thus, , so the fraction of the triangle (area ) covered by the square is . The answer is then .
Solution 5
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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