I’ve been meaning to start a problem-solving journal and write about actual problem-solving experiences. I’d been putting it off but now that it’s the summer I have no excuse… and then this blog post at Infigons, etc. inspired me!
I just picked out a possible Algebra Problem of the Week for next year that asks, “A point is chosen at random from within a circular target board. What is the probability that the point will be closer to the center of the board than to the boundary of the region?”
Infigon’s problem changes the circular dart board to a square. To which I say, “Yikes!”
My first thought is to solve the simpler problem first. I’m going to start with the circular board. My first idea is we need to find a circle that takes up half the area of the dart board. I know every point on the boundary of the circle is as close to the boundary of the dart board as every other point on the circle…
Before launching in to my drawing, I ask, “what am I trying to solve again?”
The question is, what is the probability that the point will be closer to the center of the board than the boundary of the region. That’s a different question than I was trying to solve! I was trying to find a chunk of the target that there’s a 50% chance of landing in. That would make the square problem a lot easier…
Let me go back to what I know about probability. P(Event) = # of ways event occurs / # of possible outcomes.
In an area problem we count “# of” as area (I always think of it as our best way of describing the number of points in the region, even though that’s kind of a flawed idea about infinitessimals).
So… P(being closer to the center) = Area of target closer to the center / Area of target.
How do I define closer to the center? By measuring, I guess. Pick a point, measure distance to center and distance to edge.
On a circle, that’s a piece of cake because we’d measure along the radius. So if a point is more than halfway along the radius, it’s farther from the center. If the radius of the circle is r and the center is O, the region of points closer to the center is the region centered at O with radius r/2.
The area of that is (pi*r^2)/4, which is 1/4 the area of the whole circle.
P(being closer to the center = [(pi*r^2)/4] / (pi*r^2) = 1/4.
Now how does that help me think about the harder problem? It helps me define the problem:
P(being closer to the center) = Area of target closer to the center / Area of target
The hard part of that is, what is the region of the square closer to the center? And, how do I measure the distance to the boundary? Is it to the nearest point on the boundary? How do I know.
I’m going to first check two initial ideas that seem so obvious they’re probably wrong, and then use technology to try to add some sophistication to my guess and check.
First probably wrong idea: The square that’s half as big (i.e. 1/4 the area) centered at the same point contains all the points closer to the center than the boundary. Try to prove it wrong by either finding a point in the smaller square that’s actually closer to the boundary, or finding a point outside the smaller square that’s actually closer to the center.
I figure the corner points will be possible problem spots, and sure enough, the corner points of the inner square are closer to the boundary than the center, so the region bounded by the inner square is a problem.
The inner square does have some good points, though. The midpoints of each side are halfway between the center and the nearest point on the boundary, and so will be on the edges of that mysterious region whose area I need to find.
I think, though, that there are some points that are closer to the center that might not be inside my circle. Let’s see if I can find them. I think they’ll be at the corners again… Yup! The points where the circle intersects the diagonal of the square are a LOT closer to the center than the boundary. They are in the region but not on the boundary of the region. The boundary of the region is where the distances are EQUAL from the center to the boundaries.
I’ve learned that to find out what this mysterious region looks like I need to find all the points whose distance to the center is equal to the distance to the boundary. I’ve been measuring the distance to the center along the diameter of a circle, and the distance to the boundary along a line perpendicular to the boundary.
I learned that the area of the region will be somewhere between (s^2)/4 (where s is the length of the side of the target) and (pi*s^2)/16.
There are 2 ways to go from here that I can see. One is, I know how to use Geometer’s Sketchpad to color in the parts of the square depending on the difference between the distance from the boundary and the center. So I can visualize the region whose area I want to find. The other is, to start doing some calculations and try to figure out where the boundary points are on the diagonals.
I’m in the mood for calculations first. I want to find more hard facts about this boundary.
Let R be a point on the diagonal of the target. The distance, d, from R to the center O, is easy to measure, just measure along the diagonal. It’s the radius of the circle OR. How do we measure to the nearest boundary point? It’s the horizontal or vertical distance from the point to the boundary.
To make this easier I’m going to Change the Representation: think of the square as being 1s by 1s and centered at the origin. If we put it on a graph with units of s, the square is bounded by points like (0,1), (1,1), (1,0), (1,-1), (0,-1), (-1,-1), (-1,0), (-1,1). P is on the line y = x (could be on y = -x, too, but w.l.o.g. let’s stick to y = x). Distance from P to the origin is the hypotenuse of the triangle with legs x and y. But since P is on y = x, then it’s the hypotenuse of the isosceles right triangle with leg x, and so the hypotenuse is x*sqrt(2). The horizontal distance to the boundary is 1 – x, and the vertical distance is 1 – y, but since y = x, it’s 1 – x. So P is on the boundary of our mysterious region when 1 – x = x*sqrt(2).
Doing a little algebra yields:
1 – x = x*sqrt(2)
1 = x + x*sqrt(2)
1 = x(1 + sqrt(2))
x = 1/(1 + sqrt(2))
x = 1/(1 + sqrt(2)) * (1 – sqrt(2))/(1 – sqrt(2))
x = (1 – sqrt(2))/(1 – 2)
x = (1 – sqrt(2)) / -1
x = sqrt(2) – 1
Now I can add 4 more points to my boundary. I wonder how to make sense of the quantity sqrt(2) – 1, geometrically. I also wonder how the heck to find the area of this odd shape…
Ok, now I want to use Sketchpad’s technology to see what the shape looks like around the points I’ve discovered. I used parametric coloring. I measured the distance to the nearest boundary and to the center for a point. I found the difference between those distances (distance to center – distance to boundary). When that distance went to 0 or less (i.e. closer to the center) the point would be purple. As the distance got to be greater than 1, the point went through the other colors of the rainbow. I traced the point as it moved and got:
So it does seem like the green points I find do define my region pretty well. Also… The region seems symmetrical in all sorts of ways, so that if I found the area of the region between y = x and the x axis, I could multiply it by 8 to find the area of the whole region.
The only tools I know for finding the area of a region like that are calculus tools. I’m pretty rusty on those but I know some calculations I could start to use to define my regions.
For the parametric coloring I was using sqrt(x^2 + y^2) and for the distance to the boundary I was using either 1 – x or 1 – y, whichever was smaller. In the region below y = x, we know x > y so 1 – x < 1 – y. So I should use 1 – x in that region.
I am looking for the area of the region bounded by y = 0, y = x, and sqrt(x^2 + y^2) = 1 – x. The answer to my question will be 8 times that area.
Wow, that’s a much more specific math task than “A point is chosen at random from within a square target board. What is the probability that the point will be closer to the center of the board than to the boundary of the region?” All the work prior to this has been “Understanding the Problem” and turning it into a question I could use increasingly specific mathematical tools on. I used Guess and Check, Solve a Simpler Problem, and Change the Representation a lot.
OK, back to calculus. First let’s get sqrt(x^2 + y^2) = 1 – x into y = form because I want to graph it on my picture and make sure I have the region right before I try to remember how to calculate the area of said region…
sqrt(x^2 + y^2) = 1 – x
x^2 + y^2 = 1 – 2x + x^2 [squared both sides]
y^2 = 1 – 2x
y = sqrt(1 – 2x)
Yeah, that looks right! Hold on a moment while I outline the whole region.
Now, to do some calculus. I think I’ll use my trusty Wolfram|Alpha. Because once you have a tidy little calculation to do, the part that’s fun for human calculation ends, in my opinion.
That’s approximately 0.876, which is about an 88% chance of hitting closer to the center. Is that between the upper and lower bound I originally calculated? It’s more than .25 and also more than pi/16… Hmm… that seems like a big problem since the region is smaller than the square, visually.
Oh! After much confusion I finally realized that when I defined the square to go from -1 to 1, I gave is a side length of 2 and an area of 4. Not to worry, I can still find the percentage of the area that’s in the special region: it’s about .219 or 22%. Solidly between 25% and 20% (which is pi/16).
Phew, checking your work is really important. Not all the Wolfram|Alpha stuff, but the way the situation was set up. Though I did have an error I didn’t even tell you about when I copied the wrong number from Wolfram|Alpha into the Google calculator. Caught that because the area I was getting was less than 20% and I could tell the region should take up more space than the circle did…
So, final answer: about 22%
Now, more wondering: If the area closer to the center is 25% of a circle and 22% of a square, is it less than 22% of an equilateral triangle? More than 22% of a pentagon? Will P(being closer to the center of an n sided figure) approach 25% as n approaches infinity?
And, since 1 +/- 2x seemed to be a key quantity here, does the quantity have a geometric analog? Is there a non-calculus way to solve this problem?