# 2008 iTest Problems/Problem 66

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## 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 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100
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