2018 AMC 10A Problems/Problem 23
- The following problem is from both the 2018 AMC 10A #23 and 2018 AMC 12A #17, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Area Addition)
- 3 Solution 2 (Area Addition)
- 4 Solution 3 (Similar Triangles)
- 5 Solution 4 (Similar Triangles)
- 6 Solution 5 (Similar Triangles)
- 7 Solution 6 (Similar Triangles)
- 8 Solution 7 (Coordinate Geometry)
- 9 Solution 8 (Coordinate Geometry)
- 10 Video Solution by Richard Rusczyk
- 11 See Also
Problem
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths and 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 units. What fraction of the field is planted?
Solution 1 (Area Addition)
Note that the hypotenuse of the field is and the area of the field is Let be the side-length of square
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below: Let the brackets denote areas. By area addition, we set up an equation for from which Therefore, the answer is ~MRENTHUSIASM
Solution 2 (Area Addition)
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 3 (Similar Triangles)
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 two smaller similar triangles that share a side of length . Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is . Now, let's extend this larger similar right triangle to the left until it hits the side of length . 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 , so So, the area of the square is .
Now comes the easy part--finding the ratio of the areas: .
Solution 4 (Similar Triangles)
On the diagram above, find two smaller triangles similar to the large one with side lengths , , and ; consequently, the segments with length and .
With being the side length of the square, we need to 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.
Solution 5 (Similar Triangles)
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 6 (Similar Triangles)
Let . is a triangle so and . . We know that is from the problem so . is also a triangle with . We now have . Solving this equation, we get that so the area of is . The area of the triangle is so the fraction of field that is unplanted is . Thus, the fraction of the field that is planted is .
~Heavytoothpaste
Solution 7 (Coordinate Geometry)
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 Solving this, we get . Obviously , so , and from here, the rest of the solution follows to get .
Solution 8 (Coordinate Geometry)
Let the right angle be at , the point be the far edge of the unplanted square and the hypotenuse be the line . Since the line from to the hypotenuse is the shortest possible distance, we know this line, call it line , is perpendicular to the hypotenuse and therefore has a slope of .
Since we know , we can see that the line rises by and moves to the right by to meet the hypotenuse. (Let and the rise be and the run be and then solve.) Therefore, line intersects the hypotenuse at the point . Plugging into the equation for the hypotenuse we have , and after a bit of computation we get .
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=p9npzq4FY_Y
~ dolphin7
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.