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# 1  Introduction

It is well known that group theory and Galois theory provided a thorough solution to the famous problem of determining which polynomial equations have general solutions involving only radicals and arithmetic operations. The answer is it’s not possible when the degree of the polynomial is higher than 4.

Most writings on this topic focused on this famous impossibility result. But Galois theory also gives us a roadmap to finding the solutions for cubic (degree 3) and quartic (degree 4) polynomial equations, which are quite non-trivial. I find such exercises very illuminating for understanding Galois theory and group theory.

I have always been fascinated by this topic, these notes are for crystalizing my own understandings and preserving some memory of a quite challenging and enjoyable intellectual journey, while I am still fortunate enough to be able to get my mind around such a beautiful thing. If anyone else finds it useful, that will be a bonus.

If you are interested, here is the full article: Solving Cubic and Quartic Equations Using Galois Theory

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## 偶然记得的事：顾城

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## 偶然记得的事：北岛

(欢迎看看我的博客：https://tongpeng.wordpress.com/。国内如果打不开可以去 http://blog.sina.com.cn/pengtong2016

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## The Distribution of Brownian Motion Barrier-Hitting (Passage) Time

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Given a Brownian motion with constant drift and a constant barrier, when does the Brownian motion hit the barrier? Since the barrier hitting time is random, the best we can know is the probability distribution function (PDF) of the barrier hitting time. This problem has many applications in probability theory and finance. I have learned several methods of solving this problem, I find it very instructive to compare different methods and see how techniques in one method get recast in another method from a different and often surprising angle.

If you are interested, take a look at the attached PDF file:  Brownian-Passage-Time

### Contents:

1. Introduction
2. No Drift Case
3. With Drift Case
4. Girsanov’s Theorem
5. Forward Kolmogorov Equation
6. Martingale (Optional) Stopping Theorem

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## A Elementary Solution To The Castillon’s Problem

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The Castillon’s Problem is the following planar geometry problem:

(1) Given a circle and 3 points A, B, C (can be either inside or outside the circle), construct a triangle PQR inscribed to the circle with 3 sides passing through A, B, C.

I first saw this problem when I was in middle school, it was from an old book of my father’s published perhaps in the 1940s in China (许纯舫初等几何四种). It was an exercise problem in a chapter dealing with circle, without mentioning the Castillon name. Like all good old textbooks, no solution was given, but in the text of that chapter, an example was given for the 4-point version of the problem:

(2) Given a circle and 4 points A, B, C, D (can be either inside or outside the circle), construct a quadrangle PQRS inscribed to the circle with 4 sides passing through A, B, C, D.

The solution was relatively simple: First, let’s recall some basic results about circle. By the Intersecting Chord Theorem(aka Power of a Point Theorem), for any chord PQ passing through A, product of the length of the two segments AP and AQ equals to A’s Power on the circle:

AP*AQ = OA2 – r2

Where r is the radius of the circle. Here the sign convention is: if AP and AQ are in different directions (when A is inside the circle), AP*AQ is negative; if AP and AQ are in the same direction (when A is outside the circle), AP*AQ is positive.

Now solution to (2):

X is the intersection of AB and CD. Construct point B’ on line AB such that

AB*AB’ = OA2 – r2

Since AB*AB’ = AP*AQ, the 4 points (B, P, B’ Q) are on a circle, so angles ∠AQB=∠AB’P. Similarly, ∠DSC=∠DC’P. Since the 4 points (P, Q, R, S) are on a circle, ∠AQB+∠DSC=180 => ∠AB’P+∠DC’P=180 => ∠XB’P+∠XC’P=180 => (X, B’, P, C’) are on a circle. Since X, B’, C’ are all constructible from given points A, B, C, D, point P can be found by constructing the circle passing through (X, B’, C’), then P is the intersection of this circle and circle O. So in general there can be up to two solutions.

QED (2)

Seeing this solution, I thought the 3-point version could be similarly solved. But it turned out the 3-point version, i.e. the Castillon’s Problem, was much harder. I spent many hours on it and asked people I could find who cared about such math problems, but could not find a solution. Using analytical geometry or complex number methods, I could calculate a solution, but I could not translate it back to a planar geometric solution.

Many years later, one day I was aimlessly browsing books in a Barnes & Noble, and saw Heinrich Dorrie’s classic book “100 Great Problems of Elementary Mathematics”. Problem 29 was the Castillon’s problem. Dorrie gave two solutions. The first one was a planar geometric solution, but as Dorrie mentioned, although not very long, it was intricate and not easy to see or follow. The second solution in the book was based on projective geometry. It was very elegant, and worked for any number of points. Today, when I search Castillon’s Problem on the Internet, solutions I can find are all similar to the projective geometry solution in Dorrie’s book. This solution is perhaps indeed the best solution to the Castillon’s Problem, but it requires projective geometry. I have always wondered if there is a relatively straightforward elementary planar geometric solution, especially since the 4-point version is not very difficult, can the 3-point version be “reduced” to the 4-point version?

Recently I stumbled onto such a solution. To develop this solution, it is convenient to recall the notion of Polar. As shown in the figure below, given a circle O with radius r, a point A (either inside or outside the circle) and line IJ, A’ is the intersection of lines OA and IJ, if OA⊥IJ and OA*OA’=r2, then line IJ is called the Polar of point A (relative to the circle O). The basic property of Polar is:

(3) If the Polar line of a point A (IJ) passes through point I, then the Polar line of point I (I’J) passes through Point A.

This is easy to prove: Draw line I’J passing through A and perpendicular to OI. ∠IA’A=∠II’A=90 => (I, A’, A, I’) are on a circle => OA * OA’=OI’ * OI (Intersecting Chord Theorem). IJ is the Polar of A => OA * OA’ = r=> OI’ * OI=r2 =>  I’J is the Polar of I.

QED (3)

Based on (3), since the Polar of I(AJ) passes through J, the Polar of J also passes through I, actual it is AI. This shows that points I and J are “symmetric” relative to circle O and point A: I is on the Polar of A, and the Polar of I passes through J; J is on the Polar of A, and the Polar of J passes through I. So any results for I also has a correspondence for J.

Another useful property is the following. Still refer to the figure above:

(4) Draw any line passing through point I and intersecting circle O at points S and P, then IS * IP = IA’ * IJ, and (S, P, J, A’) are on a circle.

Proof:

∠OA’J =∠JI’O = 90 => (A’, J, O, I’) are on a circle, so

IA’ * IJ = II’ * IO = (IO – I’O) * IO = IO2 – I’O*IO = IO2 – r2

On the other hand, by Intersecting Chord Theorem again, IS*IP =  IO2 – r2, so IS * IP =  IA’ * IJ, and (S, P, J, A’) are on a circle.

QED (4)

(3) and (4) above are fairly standard results. The key to my solution is the following lemma. It is an interesting result in its own right.

(5) As in the figure for (4) above, given a circle O with radius r and a point A (either inside or outside the circle), draw the Polar line of A, pick any point I on the Polar. Draw the Polar line of I, let point J be the intersection of the Polar of A and the Polar of I. As mentioned above after (3), similarly, I is the intersection of the Polar of A and the Polar of J. Now refer to the figure below. Pick any point S on circle O, extend IS to intersect circle O (again) at P. Similarly, extend JS to intersect circle O at Q. Then, Line PQ must pass through point A. Proof:

Call the intersection of lines PQ and OA A”, we want to prove that A”=A. Based on (4) above, (S, P, J, A’) are on a circle => ∠PSJ=∠PA’J. Similarly, ∠QSI=∠QA’I. It’s a basic property of circle that ∠PSJ=∠QSI=arc(PSQ)/2 => ∠PA’Q = 180 – (∠PA’J+∠QA’I) = 180-arc(PSQ). On the other hand, ∠POQ=arc(PSQ) => ∠POQ + ∠PA’Q = 180 => (P, O, Q, A’) are on a circle => ∠PA’O=∠PQO=∠QPO => triangles (OA’P) and (OPA”) are similar => OA’ * OA” = OP * OP = r2. Since IJ is the Polar line of point A, we also have OA’ * OA = r2, so A=A”.

QED (5)

How does this lemma help us solve the Castillon’s Problem? It lets us transform the Castillon’s problem involving 3 points to the 4-point version already solved in (2)!

Refer to the figure above. To repeat, the Castillon’s problem asks: Given a circle O and 3 points A, B, C, construct triangle P,Q, R inscribed to the circle O with 3 sides passing through A, B, C. We construct points I, J as in (5) above. To repeat, the process is: draw the Polar line of A, pick any point I on the Polar of A; draw the Polar line of I, J is the intersection of the Polar of A and the Polar of I. Now, using the procedure described in the solution of (2) above, we construct quadrangle RPSQ inscribed to circle O with 4 sides passing through 4 points C, I, J, B (again, the points can be either inside or outside the circle O). Based on lemma (5) above, PQ passes through A, so PQR is a triangle inscribed to circle O, with 3 sides passing through A, B, C.

QED (1) Castillon’s Problem

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