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2008 iTest Problems/Problem 66

Problem

Michael draws $\triangle ABC$ in the sand such that $\angle ACB=90^\circ$ and $\angle CBA=15^\circ$. He then picks a point at random from within the triangle and labels it point $M$. Next, he draws a segment from $A$ to $BC$ that passes through $M$, hitting $BC$ at a point he labels $D$. Just then, a wave washes over his work, so Michael redraws the exact same diagram farther from the water, labeling all the points the same way as before. If hypotenuse $AB$ is $4$ feet in length, let $p$ be the probability that the number of feet in the length of $AD$ is less than $2\sqrt3-2$. Compute $\lfloor1000p\rfloor$.

Solution

[asy] pair C=(0,0), B=(3.864,0), A=(0,1.035); draw(C--B--A--C); dot(C); label("C",C,SW); dot(B); label("B",B,SE); dot(A); label("A",A,NW); dot((1.035,0)); label("$D'$",(1.035,0),S); draw((1.035,0)--A); label("4",(1.932,0.518),NE); draw(anglemark(A,B,C,20)); label("$15^\circ$",(2.6,0),NW); draw((0,0.25)--(0.25,0.25)--(0.25,0)); [/asy]

Let $D'$ be a point on $BC$ such that $AD' = 2\sqrt3-2$ feet. By using the definitions of sine and cosine, $AC = \sqrt6 - \sqrt2$ feet and $BC = \sqrt6 + \sqrt2$ feet. From the Pythagorean Theorem, $D'C = \sqrt6 - \sqrt2$ feet.


If the number of feet in the length of $AD$ is less than $2\sqrt3-2,$ then point $D$ is within the line segment $AD',$ making point $M$ inside $\triangle ACD'.$ Since $[ACD'] = 4-2\sqrt3$ and $[ACB] = 2,$ the probability $p$ is equal to $2-\sqrt3.$


That means $1000p = 2000 - 1000\sqrt{3}.$ We can find $\lfloor 1000p \rfloor$ easily by using a calculator, but if we do not, note that $1732 < 1000\sqrt{3} < 1733,$ making $\lfloor 1000p \rfloor = 2000 - 1733 = \boxed{267}.$

See Also

2008 iTest (Problems)
Preceded by:
Problem 65 		Followed by:
Problem 67
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