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: Set up coordinates for the rectangle.
Place \(A=(0,0)\), \(B=(w,0)\), \(C=(w,h)\), \(D=(0,h)\), so ABCD goes around the rectangle in order. Q lies on AB, so \(Q=(q,0)\) where \(q=AQ\). P lies on AD, so \(P=(0,p)\) where \(p=AP\).

Step 2: Write BQ in terms of q.
\(BQ = w - q\). We are told \(BQ = 2\), so \(w = q + 2\).

Step 3: Write the three triangle areas.
Triangle PAQ is right angled at A, with legs \(AP=p\) and \(AQ=q\):
\[ \text{Area}(PAQ) = \frac{1}{2}pq \]
Triangle QBC is right angled at B, with legs \(QB=2\) and \(BC=h\):
\[ \text{Area}(QBC) = \frac{1}{2}(2)(h) = h \]
Triangle PCD is right angled at D, with legs \(DC=w\) and \(DP=h-p\):
\[ \text{Area}(PCD) = \frac{1}{2}w(h-p) \]

Step 4: Use the equal-area condition.
All three areas are equal. First set Area(PAQ) = Area(QBC):
\[ \frac{1}{2}pq = h \implies h = \frac{pq}{2} \]
Next set Area(PCD) = Area(QBC), using \(w = q+2\):
\[ \frac{1}{2}(q+2)(h-p) = h \]

Step 5: Substitute and simplify.
Put \(h = \frac{pq}{2}\) into the second equation:
\[ \frac{1}{2}(q+2)\left(\frac{pq}{2} - p\right) = \frac{pq}{2} \]
\[ (q+2)\cdot p\left(\frac{q}{2}-1\right) = pq \]
\[ (q+2)\cdot \frac{p(q-2)}{2} = pq \]
Since \(p \neq 0\) (P is a genuine point on AD, not at A), divide both sides by \(p\):
\[ \frac{(q+2)(q-2)}{2} = q \]
\[ q^2 - 4 = 2q \]
\[ q^2 - 2q - 4 = 0 \]

Step 6: Solve the quadratic.
\[ q = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5} \]
Since \(q = AQ\) is a length, it must be positive. \(1 - \sqrt{5}\) is negative (about \(-1.24\)), so it is rejected. Only \(q = 1+\sqrt{5}\) (about 3.24) is valid, so the value is unique, not "Not uniquely determined".

Final Answer:
\[ \boxed{AQ = 1+\sqrt{5}} \]