Sample Question 5:

Square \(ABCD\) has side length \(1\). Points \(P\), \(Q\), \(R\), and \(S\) each lie on a side of \(ABCD\) such that \(APQCRS\) is an equilateral convex hexagon with side length \(s\). What is \(s\)?

\(\textbf{(A) } \frac{\sqrt{2}}{3} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } 2 - \sqrt{2} \qquad \textbf{(D) } 1 - \frac{\sqrt{2}}{4} \qquad \textbf{(E) } \frac{2}{3}\)

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Answer Keys

Question 5: C

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Solutions

Question 5
Step 1: Observe that \(BP=BQ=DR=DS=1-s\). Hence, \(\triangle BPQ\) and \(\triangle DRS\) are congruent isosceles right triangles.

Step 2: In \(\triangle BPQ\), \(PQ=BP\sqrt2\), or \(s = (1-s)\sqrt2\), which simplifies to \(s = \sqrt2 - s\sqrt2\).

Step 3: Re-arrange and simplify \(s = \sqrt2 - s\sqrt2\) to get \(s = \frac{\sqrt2}{\sqrt2 + 1}\).

Step 4: To get rid of the denominator in the fraction, multiple the numerator and denominator with \(\sqrt2 - 1\), which simplifies to \(s = \frac{\sqrt2 (\sqrt2 - 1)}{\sqrt2 (\sqrt2 -1)} = \boxed{\textbf{(C) } 2 - \sqrt{2}}\).