Question

\(ABCD\) is a square with \(AB = 2\). \(P\) is the midpoint of \(AB\). The line through \(A\) that is perpendicular to \(DP\) meets the diagonal \(BD\) at \(Q\) and meets side \(BC\) at \(R\). Find the length of \(PR\).

• A. \(\dfrac{1}{2}\)
• B. \(\dfrac{\sqrt{3}}{2}\)
• C. \(\sqrt{2}\)
• D. \(1\)

Hint:

Try placing the square on coordinate axes with A at the origin; the perpendicularity condition then turns into a slope condition that is easy to solve.

Answer 1

The Correct Option is C

Step 1: Set up coordinates for the square.
Place the square so that \(A = (0,0)\), \(B = (2,0)\), \(C = (2,2)\) and \(D = (0,2)\); this matches a square of side \(AB = 2\) with the vertices labelled in order going around.
Since \(P\) is the midpoint of \(AB\), and \(A=(0,0)\), \(B=(2,0)\), we get \(P = (1,0)\).

Step 2: Find the slope of line \(DP\).
\(D = (0,2)\) and \(P = (1,0)\), so the slope of \(DP\) is \(\dfrac{0-2}{1-0} = -2\).

Step 3: Find the line through \(A\) that is perpendicular to \(DP\).
Two lines are perpendicular when their slopes multiply to \(-1\), so the slope needed here is the negative reciprocal of \(-2\), which is \(\dfrac{1}{2}\).
This line passes through \(A = (0,0)\), so its equation is \(y = \dfrac{1}{2}x\).

Step 4: Find where this line meets \(BC\), which is point \(R\).
Side \(BC\) runs from \(B=(2,0)\) to \(C=(2,2)\), so every point on it has \(x = 2\).
Substitute \(x=2\) into \(y = \dfrac{1}{2}x\): \(y = \dfrac{1}{2}(2) = 1\).
So \(R = (2, 1)\).

Step 5: Check the line also crosses diagonal \(BD\) at a point \(Q\), as the question describes.
Diagonal \(BD\) runs from \(B=(2,0)\) to \(D=(0,2)\), which has equation \(x + y = 2\).
Substitute \(y = \dfrac{1}{2}x\) into \(x+y=2\): \(x + \dfrac{1}{2}x = 2\), so \(\dfrac{3}{2}x = 2\), giving \(x = \dfrac{4}{3}\) and \(y = \dfrac{2}{3}\).
So \(Q = \left(\dfrac{4}{3}, \dfrac{2}{3}\right)\), which sits between \(A\) and \(R\) on the same line, matching the way the question describes the line crossing the diagonal first and then \(BC\).

Step 6: Compute \(PR\).
\(P = (1,0)\) and \(R = (2,1)\).
\[ PR = \sqrt{(2-1)^2 + (1-0)^2} = \sqrt{1+1} = \sqrt{2} \]

Step 7: Check the wrong options.
\(\tfrac{1}{2}\) and \(1\) are both smaller than the actual distance and do not match the coordinate calculation.
\(\tfrac{\sqrt{3}}{2}\) is close in size but comes from a different, incorrect triangle relation and also does not match.
Only \(\sqrt{2}\) matches the exact distance worked out from the coordinates.

Final Answer:
\(PR = \sqrt{2}\).
\[ \boxed{PR = \sqrt{2}} \]