Difference between revisions of "2007 iTest Problems/Problem 42"

Revision as of 13:33, 16 June 2018

Problem

During a movie shoot, a stuntman jumps out of a plane and parachutes to safety within a $100$ foot by $100$ foot square field, which is entirely surrounded by a wooden fence. There is a flag pole in the middle of the square field. Assuming the stuntman is equally likely to land on any point in the field, the probability that he lands closer to the fence than to the flag pole can be written in simplest terms as $\dfrac{a-b\sqrt c}d$ , where all four variables are positive integers, $c$ is a multple of no perfect square greater than $1$ , a is coprime with $d$ , and $b$ is coprime with $d$ . Find the value of $a+b+c+d$ .

Solution

$[asy] draw((0,0)--(100,0)--(50,50)--cycle); draw((100,0)--(100,100)--(0,100)--(0,0),dotted); real f(real x) { return 0.01(x-50)^2+25; } draw(graph(f,29.289,70.711),red); draw((29.289,29.289)--(50,50)--(70.711,29.289),red); draw((29.289,29.289)--(70.711,29.289)--(70.711,25)--(29.289,25)--cycle,dotted); [/asy]$

Using symmetry, the square can be divided into four look-alike parts. To find the probability of landing closer to the fence, subtract the outlined red area from the area of the larger triangle in the diagram.

Note that because the distance from one side of the fence to the flagpole is the same for one of the boundaries, part of the outlined red area is a parabola because the fence can be the directrix while the flagpole can be the focus. Let the left corner of the diagram be $(0,0)$ . Since the distance from the vertex to the focus is $25$ feet, the equation that models the parabola is $y = \frac{1}{100}(x-50)^2 + 25$ . Since one of the diagonals of the square can be modeled by the equation $y = x$ , solving the system can find distances. $x = \frac{1}{100}(x-50)^2 + 25$ $0 = \frac{x^2}{100} - 2x + 50$ $x = 100 - 50\sqrt{2}$ Since the distance from a point of intersection and the axis of symmetry is $50\sqrt{2} - 50$ feet, the distance between the two points of intersection with a parabola and a line as seen in the diagram is $100\sqrt{2} - 100$ feet. Using similar triangles, the height of the smaller triangle is $50\sqrt{2} - 50$ .

The area of the smaller triangle is $\frac{(100\sqrt{2}-100)(50\sqrt{2}-50)}{2} = \frac{-15000\sqrt{2} + 22500}{3}$ square feet. The area of the parabola section is a bit trickier to calculate, but by using calculus (or cleverly referring back to Problem 32), the area of the parabola section is $\frac{25000\sqrt{2} - 35000}{3}$ square feet. That means the total unwanted area is $\frac{10000\sqrt{2} - 12500}{3}$ square feet. Because the total area of the large triangle is $2500$ square feet, the wanted area is $2500 - \frac{10000\sqrt{2} - 12500}{3} = \frac{20000 - 10000\sqrt{2}}{3}$ . Thus, the probability of the stuntman dropping at a point closer to the fence is $\frac{8 - 4\sqrt{2}}{3}$ , so $a + b + c + d = \boxed{17}$ .

@@ Line 4: / Line 4: @@
 == Solution ==
+<asy>
+draw((0,0)--(100,0)--(50,50)--cycle);
+draw((100,0)--(100,100)--(0,100)--(0,0),dotted);
+real f(real x)
+{
+return 0.01(x-50)^2+25;
+}
+draw(graph(f,29.289,70.711),red);
+draw((29.289,29.289)--(50,50)--(70.711,29.289),red);
+draw((29.289,29.289)--(70.711,29.289)--(70.711,25)--(29.289,25)--cycle,dotted);
+</asy>
+Using [[symmetry]], the square can be divided into four look-alike parts.  To find the probability of landing closer to the fence, subtract the outlined red area from the area of the larger triangle in the diagram.
+Note that because the distance from one side of the fence to the flagpole is the same for one of the boundaries, part of the outlined red area is a [[parabola]] because the fence can be the directrix while the flagpole can be the focus.  Let the left corner of the diagram be <math>(0,0)</math>.  Since the distance from the vertex to the focus is <math>25</math> feet, the equation that models the parabola is <math>y = \frac{1}{100}(x-50)^2 + 25</math>.  Since one of the diagonals of the square can be modeled by the equation <math>y = x</math>, solving the system can find distances.
+<cmath>x = \frac{1}{100}(x-50)^2 + 25</cmath>
+<cmath>0 = \frac{x^2}{100} - 2x + 50</cmath>
+<cmath>x = 100 - 50\sqrt{2}</cmath>
+Since the distance from a point of intersection and the axis of symmetry is <math>50\sqrt{2} - 50</math> feet, the distance between the two points of intersection with a parabola and a line as seen in the diagram is <math>100\sqrt{2} - 100</math> feet.  Using similar triangles, the height of the smaller triangle is <math>50\sqrt{2} - 50</math>.
+The area of the smaller triangle is <math>\frac{(100\sqrt{2}-100)(50\sqrt{2}-50)}{2} = \frac{-15000\sqrt{2} + 22500}{3}</math> square feet.  The area of the parabola section is a bit trickier to calculate, but by using [[calculus]] (or cleverly referring back to [[2007 iTest Problems/Problem 32|Problem 32]]), the area of the parabola section is <math>\frac{25000\sqrt{2} - 35000}{3}</math> square feet.  That means the total unwanted area is <math>\frac{10000\sqrt{2} - 12500}{3}</math> square feet.  Because the total area of the large triangle is <math>2500</math> square feet, the wanted area is <math>2500 - \frac{10000\sqrt{2} - 12500}{3} = \frac{20000 - 10000\sqrt{2}}{3}</math>.  Thus, the probability of the stuntman dropping at a point closer to the fence is <math>\frac{8 - 4\sqrt{2}}{3}</math>, so <math>a + b + c + d = \boxed{17}</math>.
+==See Also==
+{{iTest box|year=2007|num-b=41|num-a=43}}
+[[Category:Intermediate Geometry Problems]]

2007 iTest (Problems, Answer Key)
Preceded by: Problem 41	Followed by: Problem 43
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Difference between revisions of "2007 iTest Problems/Problem 42"

Revision as of 13:33, 16 June 2018

Problem

Solution

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