On a circle with a radius of 1, pick two random points on the circumference. What is the average distance between the two points?

More precisely: start with a unit circle centered at (0, 0). Two points with polar coordinates (1, θ_{1}) and (1, θ_{2}) are chosen with θ_{i} distributed uniformly over [0, 2π]. What is the expected distance between the points?

(I was fussing about between putting the angle between [0, 2π) versus [0, 2π], but I don’t think it matters so the above image has [0, 2π) as the range.)

Watch the video for a solution.

**What Is The Average Distance Of Two Points On A Circle?**

Or keep reading.

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**Answer To Average Distance Of Two Points On A Circle?**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

We can simplify the problem a little bit. Wherever the two points are chosen, we can then rotate the circle so that one point is at (1, 0). Notice this doesn’t change the distance between the two points. The second point is equally likely to be anywhere, so we can take its angle to be randomly distributed from [0, 2π]. Actually, since the upper and lower semicircles are symmetric, we can just consider the upper semicircle and take angles from 0 to π. We will write this single angle as θ. Now we have two points in polar coordinates (1, 0) and (1, θ).

(I was fussing about between putting the angle between [0, π) versus [0, π], but I don’t think it matters so the above image has [0, π) as the range.)

To find the distance between the points, we can use the law of cosines, which gives:

(distance)^{2} = 1^{2} + 1^{2} – 2(1)(1)cos(θ)

(distance)^{2} = 2 – 2 cos(θ)

distance = √(2 – 2 cos(θ))

We can simplify this formula using a neat little trick of the sine half angle identity:

sin(*x*/2) = √((1 – cos(*x*)/2)

so then

2 sin(*x*/2) = √(2 – 2 cos(*x*))

Since θ is in [0, π] for which sine is positive, we can then write:

distance = |2 sin(θ/2)| = 2 sin(θ/2)

Now we need to calculate the average value of the above function. This is a straightforward calculation from the definition of the mean of a function over an interval.

The mean of a function *f*(*x*) over the interval [*a*, *b*] is 1/(*b* – *a*) times the integral of the function from *a* to *b*. For the distance function, we have the angle ranging from [0, π], so the average value is:

So the average distance is 4/π ≈ 1.27, which is about 27% larger than the radius of the circle. Pretty neat how the calculation simplifies!

Now for a much tougher challenge, calculate the mean distance between two points in a unit square.

See my video for a solution – average distance two points in a square.

**Source**

MathWorld Circle Line Picking

http://mathworld.wolfram.com/CircleLinePicking.html

**Trigonometry**

tan² θ + cot² θ in terms of cos θ

**Advanced Algebra**

If *a*, *b*, *c* are in Harmonical Progression, prove:

(*b* + *a*)/(*b* – *a*) + (*b* + *c*)/(*b* – *c*) = 2

Can you figure it out? Watch the video for a solution.

**Can You Solve These Stanford Admissions Questions (1892)?**

Or keep reading.

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**Answer To Stanford Admissions Questions From 1892**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

**Trigonometry**

tan² θ + cot² θ in terms of cos θ

This problem is straightforward. Recall:

tan θ = (sin θ)/(cos θ)

tan θ = (sin θ)/(cos θ)

sin² θ = 1 – cos² θ

So we substitute to get:

tan² θ + cot² θ

= (sin² θ)/(cos² θ) + (cos² θ)/(sin² θ)

= (1 – cos² θ)/(cos² θ) + (cos² θ)/(1 – cos² θ)

Now let’s solve the next problem.

**Advanced Algebra**

If *a*, *b*, *c* are in Harmonical Progression, prove:

(*b* + *a*)/(*b* – *a*) + (*b* + *c*)/(*b* – *c*) = 2

I admit I had to look this one up. In a harmonic progression, the denominators are in an arithmetic progression with some common difference *d*.

So if the first term is *a*, the next term would be *a*/(1 + *d*) = *b*, and the next term would be *a*/(1 + 2*d*) = *c*. So we can simplify the first fraction:

(*b* + *a*)/(*b* – *a*)

= (*a*/(1 + *d*) + *a*)/(*a*/(1 + *d*) – *a*)

= (*a* + *a*(1 + *d*))/(*a* – *a*(1 + *d*))

= (1 + (1 + *d*))/(1 – (1 + *d*))

= (2 + *d*)/(-*d*)

= -(2 + *d*)/(*d*)

Now we simplify the next fraction:

(*b* + *c*)/(*b* – *c*)

= (*a*/(1 + *d*) + *a*/(1 + 2*d*))/(*a*/(1 + *d*) – *a*/(1 + 2*d*))

= ((1 + 2*d*) + (1 + *d*))/((1 + 2*d*) – (1 + *d*))

= (2 + 3*d*)/(*d*)

We add these two results to get:

(*b* + *a*)/(*b* – *a*) + (*b* + *c*)/(*b* – *c*)

-(2 + *d*)/(*d*) + (2 + 3*d*)/(*d*)

= 2*d*/*d*

= 2

**Source**

Stanford Alumni magazine

http://alumni.stanford.edu/get/page/magazine/article/?article_id=46470

Every day, a train passes a train station along a straight line track, and the train moves at a constant speed.

Two friends, *A* and *B*, want to determine how long the train is. Lacking proper equipment, they devise the following method.

They first synchronize their walking. Both *A* and *B* walk at the same constant speed, and each step they take is the same length.

One day they line up back to back at the train station. When the front of the train reaches them, they both start walking in opposite directions.

Each person stops exactly as the back of the train passes by.

If person *A* takes 30 steps, and person *B* takes 45 steps, how long is the train, in terms of steps?

Watch the video for a solution.

**Can You Solve The Train Length Puzzle?**

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**Answer To Train Length Puzzle**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

The following diagram presents the given information of three stages: the start, after *A* moves 30 steps, and after *B* moves 45 steps. *A* is shown in orange and *B* is shown in green.

Notice the second diagram has *B* moving 30 steps as well, since *A* and *B* walk at the same rate.

We can deduce the speed of the train from the train’s and *B*‘s position in the second and third stages. Between those two stages, *B* moves 15 steps. The back of the train moves a total of 75 steps: it is 30 + 30 + 15.

Hence the train moves 75 steps for every 15 steps walked, so the train moves 75/15 = 5 times as fast.

Now we can calculate the train length from the first and second stages. Between the first and second stages, *A* moves 30 steps. Hence, the front of the train will move 30 × 5 = 150 steps to the right of the starting point (the vertical line). Then, from the vertical line, it is 30 steps to the left to the back of the train. So the total train length is 30 + 150 = 180 steps.

If they then measured 1 step (for example, 2.5 feet/step), they could then calculate the train length in feet too–180 × 2.5 = 450 feet.

And that’s it! We figured out the train length from a rather nifty method of walking.

]]>This is a false geometric proof that 90 equals 100. The mistake is subtle, so see if you can identify it. It is better presented in the video:

**“Prove” 90 = 100. Can You Spot The Mistake?**

But if you cannot watch the video, I present the fallacy in text/images below.

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**A false proof 90 = 100**

Start with quadrilateral *ABCD* where two opposite sides have equal length *x*, angle *A* is 90 degrees (a right angle), and angle *B* is 100 degrees.

Construct the perpendicular bisectors of *AB* and *CD* and mark the point they meet as *E*.

Now connect *E* to the vertices *A*, *B*, *C*, and *D*. Since *E* is on the perpendicular bisector of *AB*, it is equidistant from *A* and *B*, so *AE* = *BE* = *y*. Similarly *E* is on the perpendicular bisector of *CD* so it is equidistant from *C* and *D* so that *CE* = *DE* = *z*.

Now focus on triangles *AED* and *BEC*. Since they have 3 equal sides, they are congruent. Hence angles *CBE* and *DAE* are equal, call the measure α.

Now focus on triangles *FAE* and *FBE*. Since they have 3 equal sides, they are congruent. Hence angles *FAE* and *FBE* are equal, call the measure β.

We can then conclude:

angle *BAD* = α + β = 90 degrees

angle *ABC* = α + β = 100 degrees

Hence

90 degrees = α + β = 100 degrees

So we have proven that 90 equals 100, which is clearly absurd! But what is the mistake in this proof? It is a tricky one, but see if you can figure it out. I explain the mistake below.

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**The mistake in the proof**

Let’s make a diagram with a more accurate representation of the perpendicular bisectors.

Notice that point *E* is exterior to the quadrilateral; it is not on the interior of *ABCD*. This is where the mistake originates.

Why is this detail a big deal? Connect point *E* to the vertices of the quadrilateral.

It is still true, as before, that:

angle *CBE* = angle *DAE* = α

angle *FAE* = angle *FBE* = β

The mistake in the previous proof was saying the corresponding equal angles added up to *BAD* and *ABC*. In other words, the mistake is:

*ABC* ≠ α + β

*BAD* ≠ α + β

So the proof above did not in fact show that 90 degrees equals 100 degrees after all!

The false proof illustrates an important lesson that you cannot assume the location of a constructed point like *E*, particularly when you use a diagram that is not to scale!

The blue circle’s radius is 4 and the green circle’s radius is 2. The line is tangent to all three circles, and the three circles touch each other (they are pairwise tangent). What is the radius of the white circle?

Give it a try, and then watch the video for a solution.

**Solve For The Radius. Challenging Problem From Indonesia!**

Or keep reading.

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**Answer To Great Problem For Indonesia 15 Year Olds**

Let *r* denote the radius of the smallest circle (the white circle in the original diagram).

The key to this problem is creating three right triangles, each of which has a hypotenuse that connects the center’s of two circles:

Here’s how the diagram works out step by step. We can draw the blue triangle, and it is straightforward to see its hypotenuse as 4 + *r* and one leg as 4 – *r*.

We now can use the Pythagorean Theorem to get the other leg:

√((4 + *r*)^{2} – (4 + *r*)^{2})

= 4√*r*

We can do a similar calculation for the green triangle. Its other leg is:

√((2 + *r*)^{2} – (2 – *r*)^{2})

= 2√2(√*r*)

Now we can draw the purple triangle, which has a hypotenuse 4 + 2 = 6 and one leg 4 – 2 = 2. Its last leg is equal to the sum of the two lengths calculated above, as this leg is parallel to the legs of the blue and green triangles. So the last leg is:

4√*r* + 2√2(√*r*)

= √*r*(4 + 2√2)

Now we use the Pythagorean Theorem to get:

(√*r*(4 + 2√2))^{2} + (4 – 2)^{2} = (4 + 2)^{2}

*r* = 32/(4 + 2√2)^{2}

*r* = 12 – 8√2 ≈ 0.686

We can actually generalize the result using the same steps as above. Suppose the larger circles have radii *R*_{1} and *R*_{2}. We can again create three right triangles:

We can then do the same steps and solve the smallest circle’s radius satisfies:

1/√*r* = 1/√*R*_{1} + 1/√*R*_{2}

I can see the last formula might be useful in competition mathematics!

**Thanks to all patrons! Special thanks to:**

Shrihari Puranik

Kyle

Professor X

Share the beauty of math with the world! You can support these videos and posts at Patreon and get exclusive rewards: http://www.patreon.com/mindyourdecisions

**Websites consulted**

Descartes Theorem (the above problem is a special case)

https://en.wikipedia.org/wiki/Descartes%27_theorem

Cut The Knot Tangent Circles

https://www.cut-the-knot.org/pythagoras/TangentCirclesSangaku.shtml

Lost Math Lessons Three Kissing Circles And A Line

http://lostmathlessons.blogspot.com/2015/12/kissing-circles-three-circles-and-line.html