2006 iTest Problems/Problem 8

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Problem

The point $P$ is a point on a circle with center $O$. Perpendicular lines are drawn from $P$ to perpendicular diameters, $AB$ and $CD$, meeting them at points $Y$ and $Z$, respectively. If the diameter of the circle is $16$, what is the length of $YZ$?

$\mathrm{(A)}\,4\quad\mathrm{(B)}\,8\quad\mathrm{(C)}\,6\sqrt{3}\quad\mathrm{(D)}\,4\sqrt{3}\quad\mathrm{(E)}\,4\sqrt{2}\quad\mathrm{(F)}\,12\quad\mathrm{(G)}\,6\quad\mathrm{(H)}\,\text{none of the above}$

Solution

[asy] pair P=(-60,80), A=(-100,0), B=(100,0), C=(0,100), D=(0,-100), Y=(-60,0), Z=(0,80); draw(circle((0,0),100)); draw(A--B); draw(C--D); draw(Y--P--Z); draw(Y--Z); dot((0,0)); label("O",(0,0),SE); dot(P); label("P",P,NW); dot(A); label("A",A,W); dot(B); label("B",B,E); dot(C); label("C",C,N); dot(D); label("D",D,S); dot(Y); label("Y",Y,S); dot(Z); label("Z",Z,E); [/asy]

Since $AB \perp CD, AB \perp PY, CD \perp PZ$, we know that $PYOZ$ is a rectangle. The diagonal of a rectangle is the same length, so $YZ = PO$. Since $PO$ is the radius of the circle, the length of $YZ$ equals $\tfrac12 \cdot 16 = \boxed{\textbf{(B)}\,8}$.

See Also

2006 iTest (Problems)
Preceded by:
Problem 7
Followed by:
Problem 9
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