March 2026 Problem of the Month

A circle with area π square inches is inscribed inside a right triangle.
What are the lengths of the sides of the triangle?

Solution to the Problem:

Here is the solution:

The sides of the right triangle are 3 inches, 4 inches, and 5 inches.

A circle with area 𝜋 has radius 𝑟 = 1.

For a right triangle with legs a and b and hypotenuse 𝑐, the inradius is given by the formula
r = (a + b - c) /2
Set r = 1, and you obtain a + b - c = 2

Using the Pythagorean theorem along with the previous equation, you obtain a=3,b=4,c=5.

---

Davit Banana gets extra credit for finding more than one solution.
Davit pointed out that I did not specify integer solutions, so he sent the following message:

For Americans, the comma in the table corresponds to our decimal point.

---

Danny of Georgia and Kelly Stubblefield also get extra credit for pointing out that there is an infinite number of solutions.

---

Kamal Lohia also gets extra credit for pointing out that there is an infinite number of solutions.
Here is his excellent explanation:

Answer: 3, 4, 5 inches

Solution: Area of circle = πr² = π
=> r = 1

So, the three sides of the right triangle are: 1+x, 1+y, x+y

Using Pythagorean theorem, we can write
(1+x)² + (1+y)² = (x+y)²
=> 1+2x+1+2y = 2xy
=> xy-x-y+1 = 2
=> (x-1)(y-1) = 2
=> x,y = 3,2 or 2,3 considering only integral cases else there are infinite solutions.

---

Dragan Gonzales determiined the sides of an isosceles right triangle that satisfied the original conditions:

Assuming an isosceles right triangle; the hypotenuse is (2+2(2)^1/2) units long, and the legs are each (2+(2)^1/2) units long.

Let points a, b, and c form a right triangle with, b as the right angle, and point d be the midpoint of line ac, the hypotenuse.
Let circle x be inscribed in triangle abc and have an area of pi.
By definition of an inscribed circle, line bd intersects point x.
Let points e and f be where the circle intersects lines ab and bc respectively, thus making them perpendicular to their respective line.
Circle x has an area of pi = pi * r * r, thus a radius of 1, so lines xe and xf are 1 unit long.
This makes square xbef with side lengths of 1 unit.
Line bx now makes the right triangle bex, meaning line bx has a length of 2^.5.
Line xd is a radian so its length is 1, thus line bd has a length of 1 +2^.5.
Line bd makes triangles bcd and abd, which are isosceles right triangles with leg lengths of 1+2^.5 units.
Thus, line ac = 2(1+2^.5) = 2+2(2^.5) units, and lines ab and bc = 2^.5(1+2^.5) = 2+2^.5.

So ac = 2 + 2 √2 ≈ 4.828 and ab and bc = 2 + √2 ≈ 3.414.