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2018 AMC 10A Problem 23

2018 AMC 10A Problem 23AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 4 of 5 · about 5 min · position P21-25

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 $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?

A)
$\frac{25}{27}$
B)
$\frac{26}{27}$
C)
$\frac{73}{75}$
D)
$\frac{145}{147}$
E)
$\frac{74}{75}$

Answer

Key insight

Cut the field into the square, a base-5 height-2 triangle on the hypotenuse, and two right triangles along the legs; total area 6 forces s = 2/7.

Solution

Let the square have side $s$ and let $P$ be its corner farthest from the right angle. The nearest point of $S$ to the hypotenuse is $P$ , so the perpendicular from $P$ to the hypotenuse has length $2$ .

Cut the whole field (area $\tfrac12 \cdot 3 \cdot 4 = 6$ ) into four pieces by drawing segments from $P$ to the two acute vertices:

the square $S$ , area $s^2$ ;
the triangle with base the hypotenuse ( $5$ ) and apex $P$ , height $2$ , area $\tfrac12 \cdot 5 \cdot 2 = 5$ ;
a right triangle along the leg of length $4$ with legs $s$ and $4 - s$ , area $\tfrac12 s(4 - s)$ ;
a right triangle along the leg of length $3$ with legs $s$ and $3 - s$ , area $\tfrac12 s(3 - s)$ .

Adding:
$s^2 + 5 + \tfrac12 s(4 - s) + \tfrac12 s(3 - s) = 6.$
The $s^2$ terms cancel ( $s^2 - \tfrac12 s^2 - \tfrac12 s^2 = 0$ ), leaving $\tfrac72 s = 1$ , so $s = \tfrac27$ .

The unplanted area is $s^2 = \tfrac{4}{49}$ , so the planted fraction is
$1 - \frac{4/49}{6} = 1 - \frac{2}{147} = \frac{145}{147}.$

The answer is $\boxed{\textbf{(D)}\ \frac{145}{147}}$ .

Why this works

A "distance to a line" clue is an altitude in disguise: the segment of length $2$ is the height of a triangle whose base is the entire hypotenuse, and that triangle's area ( $5$ ) is known instantly. Comparing the total area with the sum of the pieces converts a geometric constraint into a linear equation in $s$ . Dissection by connecting the key point to the vertices is a reusable move for any "point inside a triangle at known distance from a side" problem.

Alternative approach

Coordinates: right angle at the origin, hypotenuse on the line $3x + 4y = 12$ , corner $P = (s, s)$ . Distance from $P$ to the line:
$\frac{|3s + 4s - 12|}{5} = \frac{12 - 7s}{5} = 2 \quad\Longrightarrow\quad s = \frac27.$
Then the same fraction $1 - \dfrac{4}{49 \cdot 6} = \dfrac{145}{147}$ .

The trap

Assuming the square's far corner lies on the hypotenuse, or measuring the distance 2 from the wrong point (the right-angle vertex instead of the square's outer corner).

Common mistakes

Assuming the square's far corner lies on the hypotenuse, or measuring the distance 2 from the wrong point (the right-angle vertex instead of the square's outer corner).
Computing the unplanted fraction $\tfrac{2}{147}$ and stopping, or dividing $s^2$ by $12$ (the product of the legs) instead of the area $6$ .

Techniques

Place the figure on coordinates and compute · Cut the figure into known shapes (triangles, rectangles, sectors)