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A semicircle is drawn on the side AB of a rectangle ABCD. Let P be a point on the semicircle, and let X and Y be defined as shown in the diagram.
If the value \({AX}^2+{BY}^2\) is independent of the position of P, then find \(\frac{AD}{AB}\).
A. 0.5
B. \(\frac{1}{\sqrt{2 }}\)
C. 1
D. 1.5
E. \(\sqrt{3}\)
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xCejA2XNAK-20200329_215723.jpg [ 168.22 KiB | Viewed 2923 times ]
31 Mar 2020, 21:46
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Assume AD = 'a' and AB = 'b'
Case 1. When P and A coincides
AX = 0 and BY = a
\({AX}^2+{BY}^2\) = \(a^2\)
Case 2. When P is at the midpoint of the arc
triangle PYX is similar to triangle PDC
\(\frac{PM}{PN} = \frac{YX}{DC}\)
[(a/2)]/[(a/2)+b] = YX/a
\(YX= \frac{a^2}{a+2b}\)
\(AX= BY =a/2+\frac{YX}{2} = \frac{a^2+ab}{a+2b}\)
\({AX}^2+{BY}^2\) = \(2a^2 (\frac{a+b}{a+2b})^2\)
Since \({AX}^2+{BY}^2\) is independent of position of P, it must be equal in both cases.
\(a^2 = 2a^2 (\frac{a+b}{a+2b})^2\)
\(a^2 = 2b^2\)
\(\frac{b}{a}\) = \(\frac{1}{\sqrt{2 }}\)
Case 1. When P and A coincides
AX = 0 and BY = a
\({AX}^2+{BY}^2\) = \(a^2\)
Case 2. When P is at the midpoint of the arc
triangle PYX is similar to triangle PDC
\(\frac{PM}{PN} = \frac{YX}{DC}\)
[(a/2)]/[(a/2)+b] = YX/a
\(YX= \frac{a^2}{a+2b}\)
\(AX= BY =a/2+\frac{YX}{2} = \frac{a^2+ab}{a+2b}\)
\({AX}^2+{BY}^2\) = \(2a^2 (\frac{a+b}{a+2b})^2\)
Since \({AX}^2+{BY}^2\) is independent of position of P, it must be equal in both cases.
\(a^2 = 2a^2 (\frac{a+b}{a+2b})^2\)
\(a^2 = 2b^2\)
\(\frac{b}{a}\) = \(\frac{1}{\sqrt{2 }}\)
nick1816
02 Apr 2020, 08:41
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