2018 AMC 10A Problems/Problem 23
Contents
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 1
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 .
Alternatively, once you get , you can avoid computation by noticing that there is a denominator of , so the answer must have a factor of in the denominator, which only does.
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 by ktong
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
On the diagram above, find two smaller triangles similar to the large one with side lengths , , and ; consequently, the segments with length and .
Find an expression for : using the hypotenuse, we can see that . Simplifying, , or .
A different calculation would yield , so . In other words, , while to check, . As such, , and .
Finally, we get , to finish. As a proportion of the triangle with area , the answer would be , so is correct. ∎ --anna0kear
Solution 6
Let be the side length of the square. The area of the triangle is . Connect the inside corner of the square to the three corners. Then, the area of the triangle is also . Solving gives . That makes the answer .
Solution 7 (Trig)
Let A be the bottom left corner of the triangle. Using similar triangles, we deduce that the distance from the hypotenuse to A is . Call that line a. Let B be the top right corner of the square; the shortest distance from B to the hypotenuse is 2. Call that line b. Draw a perpendicular from B down to the line connecting A to the hypotenuse. Call the intersection point C. Since line a has length , AC has length . Clearly ACB is a right triangle. Call the angle of the big triangle between 4 and 5 x. Since AB is 45 degrees from the horizontal, is 45+x. Using the addition sine formula, we deduce that . From that, we find that AB is , so the side length of the square is 2/7. Therefore the area of the square is 4/49, and the answer we want is D) 145/147.
-EZmath2006
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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