Areas and Tangent Circles

After many questions about trigonometry, calculus, and parabolas, we finally get a fresh new question about geometry. This question deals with one of the most hated figures in the world of geometry: circles.

Two circles with radii of 5 cm each are externally tangent to each other. Each of these two circles is internally tangent to a circle with a radius of 15 cm. What is the perimeter and area of the shape enclosed between the three circles?

To get the idea of what it looks like, try to draw it. You will eventually see what this whole question is about, and that will make the question much easier for you.

First, we need to get the perimeter. When you connect all the centers of the circles, you get an equilateral triangle. This happens because the side lengths are 5 + 5 and 15 - 5 (externally tangent circles and internally tangent circles, respectively). This triangle is the key to the whole question.

There is a theorem saying that the centers of two tangent circles and the point of tangency all lie on the same line. This means you can extend two sides of the equilateral triangle to create a sector of the big circle. This sector's area and perimeter are 60°/360° = 1/6 of the area and the perimeter of the circle.

Since these extensions go through the centers of the small circles as well, they split them up to 60° and 120° angles (the 60° angle is formed by the equilateral triangle, and the 120° angle is its supplement). Now we can calculate both the perimeter and the area of the shape between the three circles.

The perimeter is given by adding the three arcs which form the shape. Each small circle gives an arc of 120°, and the big circle gives an arc of 60°. Let's find these arc lengths.

Small circles:
2π * 5 * 120°/360° = 10π/3

Large circle:
2π * 15 * 60°/360° = 30π/6 = 5π

There are two small circles, so the total perimeter is:
5π + 2 * 10π/3 = 15π/3 + 20π/3 = 35π/3 cm
This equals about 36.65 cm.

As for the area, we can calculate the area of the sector of the large circle, subract the area of the equilateral triangle, and subtract the area of the two sectors of the small circles. This will give us the area of the shape we want.

Area of the big sector:
π * 15² / 6 = 225π / 6 = 37.5π

Area of the equilateral triangle:
10²√3 / 4 = 25√3

Area of a small sector:
π * 5² / 3 = 25π/3

Remember there are two small sectors, so we can now calculate:
37.5π - 25√3 - 2 * 25π/3 = 225π/6 - 25√3 - 100π/6 = 125π/6 - 25√3 cm²
This equals about 22.15 cm²

That's it for today. Hope you liked it. Visit super math tips for more cool information like that.

Yours,
Nadav

nadavs

Coordinated Circles

Today I have a very common question - how to find an equation of a circle from three points, which are a requirement for a unique circle. See how this can be done the hard way and the easy way.

Find the center, radius, and area of a circle that passes through (6, 2), (0, -6), and (7, 1).

The usual way of doing this type of problems is finding the equations of two lines between those points, finding their perpendicular bisectors, find their point of intersection (the center), and go from there. There is one very good adjective to describe this: boring. Since this blog is all about interesting math, you'll see there is a way that cuts all this line of operations up to "finding their perpendicular bisectors". Yes, removing two very annoying steps.

Since the center of the circle is equidistant from all three points, we need to find a point which is equidistant from the first two points, then find a point which is equidistant from the second and third point, and then see if they match. If they do, the point we found is equidistant from the first and third point (by the transitive property).

So, let's equate the distance between (6, 2) and (0, -6) (the generic point is (x, y)):
√((x - 6)² + (y - 2)²) = √((x - 0)² + (y + 6)²)

Square both sides:
(x - 6)² + (y - 2)² = x² + (y + 6)²

Open parentheses and add like terms:
x² - 12x + 36 + y² - 4y + 4 = x² + y² + 12y + 36
-12x - 4y + 4 = 12y
-12x - 16y + 4 = 0
-16y = 12x - 4
y = -3x/4 + 1/4

We didn't get a point, we got a line. This line is the perpendicular bisector of the segment between the two points, because the geometric location of all the points equidistant from two points is a line. This line is the perpendicular bisector of the segment connecting the two points. Now you know how to find a perpendicular bisector of two points in one very easy step.

Let's do the same thing to the second pair of points:
√((x - 0)² + (y + 6)²) = √((x - 7)² + (y - 1)²)
x² + (y + 6)² = (x - 7)² + (y - 1)²
x² + y² + 12y + 36 = x² - 14x + 49 + y² - 2y + 1
12y + 36 = -14x + 49 - 2y + 1
14y = -14x + 14
y = -x + 1

Now we have two sets of points that are equidistant from a given set of points. All we need now is to equate those lines and find the point that is equidistant from all three points:
y = -3x/4 + 1/4
y = -x + 1

Equate the y's:
-3x/4 + 1/4 = -x + 1
x/4 = 3/4
x = 3

So:
y = -3 + 1
y = -2

The point which is equidistant from all three points and thus the center of the circle is (3, -2).

Now everything becomes really easy. To find the radius of the circle, we need to find the distance from the center to one of the points. Let's say to (7, 1):
r = √((7 - 3)² + (1 - (-2))²)
r = √(4² + 3²)
r = √25
r = 5

Using the formula for the area of a circle, we finally answer the question completely with the answer:
S = 5² * π = 25π

Problem solved.
Nadav

nadavs

Constructing a Square Root

Today's question is not from Yahoo Answers, but rather from my head. Well, not exactly from my head. I read about the possibility to construct a square root of a number, but I didn't find any instructions about how to do it. So, here is what I found out.

There is a line of length x. Construct a line of length sqrt(x)

If you remember anything about averages, there are two common types. There is a simple arithmetic average, simply adding two numbers and halving the result. There is also a geometric average, which is multiplying two numbers and taking the square root of the result.

This average is called a geometric average because it happens in, surprisingly, geometry. In a right triangle, the altitude from the 90 degree angle to the hypotenuse is equal to the square root of the two parts of the hypotenuse created from the altitude. If the altitude creates segments of length x and y, then the altitude's length is sqrt(xy). We can use that to find sqrt(x). All we have to do is make y = 1.

So to construct this segment of length sqrt(x) we construct a segment with length x and a segment with length 1 (the 1 is relative to x. If x is 5, the segment of length 1 must be 5 times shorter). Now we created the hypotenuse. Now construct the altitude to the hypotenuse from the point the two segments are joined.

We are in a little trouble now (a trouble that made me think for almost 30 minutes). We need to find the exact point on the altitude that will create a right angle with the endpoints of the hypotenuse. However, we have no way of knowing where is that point, since we construct a square root. Unless it's a square root of a perfect square, we cannot count units of 1 and create the right triangle from there.

After almost 30 minutes of thinking, it hit me. When a right triangle is inscribed inside a circle, the hypotenuse is a diameter. Now we can create an exact right triangle and an exact square root. Find the midpoint of the hypotenuse and create a circle around it.

Now it's easy. Continue the altitude until it intersects the circle. Then you can complete the triangle and take the altitude. Square root found.

Hope you liked it. Now go test it - find the square roots of numbers like 4 and 9 and see if you get 2 and 3. If you do, you construct well.

Yours,
Nadav

nadavs

Circular Coordinate Geometry

Today I have a question involving some coordinate geometry, or analytic geometry. It also involves some circles. It's quite easy, but somewhat long. Here it comes.

What is the sum of the radii of all circles going through the points (1, 9) and (8, 8) and also tangent to the x-axis?

First, you should know that if a circle is tangent to the x-axis, its radius is the y-coordinate of its center. Let's call the center (x, y), so its radius is y.

Since it's a circle, the distance between the center and each of the points must be y. Let's use the distance formula to show that:

y = sqrt((x - 1)² + (y - 9)²)
y = sqrt((x - 8)² + (y - 8)²)

Square and open parentheses:
y² = x² - 2x + 1 + y² - 18y + 81
y² = x² - 16x + 64 + y² - 16y + 64

y^2 cancels on all equations (I also moved the y terms):
18y = x² - 2x + 82 /*8
16y = x² - 16x + 128 /*9

144y = 8x² - 16x + 656
144y = 9x² - 144x + 1152

9x² - 144x + 1152 = 8x² - 16x + 656
x² - 128x + 496 = 0
(x - 4)(x - 124) = 0
x = 4, 124

Now let's see what's y (the radius) for each x:
16y = 4² - 16*4 + 128 = 80
y = 5

When x = 124:
16y = 124² - 16*124 + 128 = 13520
y = 845

So the two circles have radii of 5 and 845, so the sum of the radii is 850.

Hope you liked it.
Nadav

nadavs

Geometric Construction and the Sinus Function

Today I have a fantastic question from Yahoo Answers about geometric construction. The idea of geometric construction is to use a compass and a ruler (with no measurements) to construct different geometric shapes and figures.

Notice: to answer this question you need to know basic constructions: copying a segment, creating a segment X times larger than another, find a perpendicular bisector, and copy an angle. Without those you'll be lost.

Here is the question:

A triangle has sides a, b, and c. The ratio a/b = 7/4. You are given side c and the radius of the circumcircle, r. Construct the triangle.

This may seem simple, but it's a little more sophisticated than that. To construct that triangle, we need the law of sines.

The law of sines says that in a triangle, a/sin A = b/sin B = c/sin C = 2R (side a and angle A are opposite, R is the radius of the circumcircle). Let's play with this law a little:
a/sin A = b/sin B
That means:
a/b = sin A / sin B

In our triangle, a/b = 7/4. So all we need to do is find two angle whose ratio of sines is 7/4. So how do we do that?

The definition of sine A is the y-coordinate on the unit circle with the angle measure of A (read here about the unit circle). First, construct two segments, a' and b' in a way that a'/b' = 7/4. Now create a circle with a radius of more than a' and two perpendicular axes that intersect at the center of the circle. Position a' in the circle in such a way that it's perpendicular to the x-axis and touches the circle in one point (yet not tangent to it). Construct the line from the origin to the point of intersection of a' and the circle. Call the angle between the x-axis and that line angle A. Do the same thing for b' and angle B.

Since we used the definition of sine, we now have two angles with a sine ratio of 7/4. Now we need to create the triangle we want.

First, construct segment c. Since the circumcenter, the center of the circumcenter, is on the point of intersection of all perpendicular bisectors, construct the perpendicular bisector of segment c. Now, since the radius is given, use the end point of segment c and the perpendicular bisector to find the circumcenter and draw the circumcircle.

Now, when segment c is inside the circle, copy angle A to one of its sides and angle B to the other side. Complete the triangle, and you are done.

Feel free to send in more questions!
Nadav

nadavs

Geometry and Road Design

Today I have a very interesting question involving geometry. Its solution is so simple that it will make you think "How didn't I think about it?!" when you see the answer. Here it is, try to solve yourself first (as I said, it's easy):

Two towns, town 1 and town 2, are located to the south of road A going in east-west direction. No roads are connecting the towns to each other or to road A. The people of both towns decided to build two roads: one from each town to road A, so they could travel between the towns. What is the shortest route to build these roads?

Well, the shortest route is to build these roads from town to town, but the people of the towns want a connection to road A, so let's give them that.

To solve this question, let's shift our thinking a little bit: let's imagine that town 1 moves to the north of road A, but still keeps the same distance from it (if it was 100 miles to the south, now it's 100 miles to the north).

The shortest distance between two points is on the straight line between them, and so is the road between the towns now. Notice that this road intersects road A.

Now move town 1 back to where it was, and create the road from town 1 to where the roads between the towns intersected road A. It's the same distance as when town 1 was at north, so it must be the shortest distance. Problem solved!

This principle is also known in physics in the field of optics and mirrors. Who said math has no use in real life?

Nadav

nadavs