2018 AMC 10A Problems/Problem 23
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: