Sample Question 17:

In rectangle \(PQRS\), \(PQ=8\) and \(QR=6\). Points \(A\) and \(B\) lie on \(\overline{PQ}\), points \(C\) and \(D\) lie on \(\overline{QR}\), points \(E\) and \(F\) lie on \(\overline{RS}\), and points \(G\) and \(H\) lie on \(\overline{SP}\) so that \(AP=BQ<4\) and the convex octagon \(ABCDEFGH\) is equilateral. The length of a side of this octagon can be expressed in the form \(k+m\sqrt{n}\), where \(k\), \(m\), and \(n\) are integers and \(n\) is not divisible by the square of any prime. What is \(k+m+n\)?

\(\textbf{(A) }1 \qquad \textbf{(B) }7 \qquad \textbf{(C) }21 \qquad \textbf{(D) }92 \qquad \textbf{(E) }106 \qquad\)

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Answer Keys

Question 17: B

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Solutions

Question 17
Step 1: Assign variable to the unknowns: Let \(AP=BQ=x\). This means that \(AB=8-2x\).

Step 2: Consider the side \(CD\), since the octagon is equilateral, its length would also be \(8-2x\). To find \(CD\), we subtract the length of rectangle sides \(PQ\) and \(RS\) by 2 times \(x\) (the length of \(AP\) and \(BQ\)). This implies that \(QC=DR=x-1\).

Step 3: Apply the Pythagorean theorem: We know that \(\overline{AC} = \overline{DR}\) in length, so we can create a right triangle \(\triangle ABC\) where \(BC\) is the hypotenuse. Consequently, we can apply the Pythagorean theorem on it. Thus, we have \(x^2 + (x-1)^2 = (8-2x)^2\).

Step 4: Simplify the equation: The equation derived in the previous step simplifies to \(2x^2 -30x +63 =0\). This equation can be solved for \(x\), yielding \(x=\dfrac{15-3\sqrt{11}}{2}\).

Step 5: Substitute \(x\) to find the length of the side: The length of the side of the octagon would be the distance \(AB=8-2x\). Thus substituting we get \(8-(15-3\sqrt{11})=3\sqrt{11}-7\).

Step 6: Compute for \(k+m+n\): The length of the side in form \(k+m\sqrt{n}\) gives \(k=-7\), \(m=3\), and \(n=11\), therefore, \(k+m+n=-7+3+11 = 7\).

Hence, the answer is \(\boxed{\text{(B)}~7}\).