Question

Let PQRS be a square inscribed in the triangle with vertices A(0, 0), B(3, 0) and C(2, 1). Given that P, Q are on the side AB, R on side BC and S on side AC. The centre of square PQRS is

• A. $\left(\frac{13}{8},\frac{8}{3}\right)$
• B. $\left(\frac{15}{8},\frac{3}{8}\right)$
• C. $\left(\frac{15}{8},\frac{8}{3}\right)$
• D. $\left(\frac{14}{3},\frac{8}{3}\right)$

Answer 1

Answer: (B)

$\left(\frac{15}{8},\frac{3}{8}\right)$

Answer 2

Solution:

1. Assume the square has side length s. Let P = (a, 0) and Q = (a+s, 0) lie on AB (x-axis). Then, since PQ is horizontal, the square’s next vertices are R = (a+s, s) and S = (a, s).

2. Since S lies on AC (from A(0, 0) to C(2, 1)), whose equation is y = (1/2)x, we set: s = (1/2)·a ⟹ a = 2s.

3. Since R lies on BC. BC connects B(3, 0) and C(2, 1); its equation is found using slope m = (1 - 0)/(2 - 3) = –1, so: y = 3 - x. For R = (a+s, s), we have: s = 3 - (a+s) ⟹ s = 3 - a - s ⟹ 2s = 3 - a. Substitute a = 2s: 2s = 3 - 2s ⟹ 4s = 3 ⟹ s = 3/4. Then, a = 2s = 3/2.

4. So the square’s vertices are: P = (3/2, 0), Q = (3/2 + 3/4, 0) = (9/4, 0), R = (9/4, 3/4), S = (3/2, 3/4).

5. The center of the square is the midpoint of the diagonal from P to R: Center = ( (3/2 + 9/4)/2 , (0 + 3/4)/2 ) = ( (6/4 + 9/4)/2, (3/4)/2 ) = (15/8, 3/8).

Explanation (Minimal):

• Let P = (a, 0), Q = (a+s, 0), R = (a+s, s), S = (a, s) with S on AC: s = (1/2)a ⟹ a = 2s.
• For R on BC (line: y = 3 - x): s = 3 - (a+s) ⟹ 2s = 3 - a ⟹ s = 3/4, a = 3/2.
• Center of square = (P + R)/2 = (15/8, 3/8).