Question

ABCD is a rectangle. The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same areas and BQ = 2 then AQ = ?

• A. \(1+\sqrt{5}\)
• B. \(1-\sqrt{5}\)
• C. \(\sqrt{7}\)
• D. \(2\sqrt{7}\)

Hint:

Set up right triangles at three corners of the rectangle, express their areas using AQ, BQ, AP and AD, then equate them to form and solve a quadratic in AQ.

Answer 1

The Correct Option is A

Step 1: Name the unknowns directly on the rectangle.
Let the rectangle have sides AB and AD. Let $AQ = x$ (so Q splits AB), and since $BQ = 2$, the full side $AB = x+2$. Let $AD = h$ and let $AP = y$, so $PD = h-y$.

Step 2: Express each corner triangle's area using the rectangle's own sides.
Triangle QBC sits in the corner at B, with legs $QB=2$ and $BC=AD=h$:
\[ [QBC] = \frac12 \cdot 2 \cdot h = h \]
Triangle PAQ sits in the corner at A, with legs $AP=y$ and $AQ=x$:
\[ [PAQ] = \frac12 xy \]
Triangle PCD sits in the corner at D, with legs $DC=AB=x+2$ and $DP=h-y$:
\[ [PCD] = \frac12 (x+2)(h-y) \]

Step 3: Equate [PAQ] and [QBC] to link h and the product xy.
\[ \frac12 xy = h \ \Rightarrow\ xy = 2h \]

Step 4: Equate [PCD] and [QBC], then remove h using Step 3.
\[ \frac12 (x+2)(h-y) = h \ \Rightarrow\ (x+2)(h-y) = 2h \]
Replace $h$ with $\frac{xy}{2}$ throughout:
\[ (x+2)\left(\frac{xy}{2} - y\right) = 2\cdot\frac{xy}{2} \]
\[ (x+2)\cdot y\left(\frac{x}{2}-1\right) = xy \]
Cancel $y$ (a genuine side length, so $y \neq 0$):
\[ (x+2)\left(\frac{x-2}{2}\right) = x \]
\[ (x+2)(x-2) = 2x \]
\[ x^2 - 4 = 2x \]
\[ x^2 - 2x - 4 = 0 \]

Step 5: Solve and pick the valid root.
By the quadratic formula, $x = 1 \pm \sqrt5$. A length cannot be negative, and $1-\sqrt5 \approx -1.24$ is negative, so it is thrown out. That leaves a single valid answer, so AQ is in fact uniquely determined by the given data.

Final Answer:
\[ \boxed{AQ = 1 + \sqrt5} \]