# Chaitanya's Random Pages

## March 19, 2017

### Two similar geometry problems based on perpendiculars to cevians

Filed under: mathematics — ckrao @ 7:18 am

In this post I wanted to show a couple of similar problems that can be proved using some ideas from projective geometry.

The first problem I found via the Romantics of Geometry Facebook group: let $M$ be the point of tangency of the incircle of $\triangle ABC$ with $BC$ and let $E$ be the foot of the perpendicular from the incentre $X$ of the $\triangle ABC$ to $AM$. Then show $EM$ bisects $\angle BEC$.

The second problem is motivated by the above and problem 2 of the 2008 USAMO: this time let $AM$ be a symmedian of $ABC$ and $E$ be the foot of the perpendicular from the circumcentre $X$ of $\triangle ABC$ to $AM$. Then show that $EM$ bisects $\angle BEC$.

Here is a solution to the first problem inspired bythat of Vaggelis Stamatiadis. Let the line through the other two points of tangency $P, Q$ of the incircle with $ABC$ intersect line $BC$ at the point $N$ as shown below. Note that since $AP$ and $AQ$ are tangents to the circle, line $NPQ$ is the polar of $A$ with respect to the incircle.

Since $N$ is on the polar of $A$, by La Hire’s theorem, $A$ is on the polar of $N$. The polar of $N$ also passes through $M$ (as $NM$ is a tangent to the circle at $M$). We conclude that the polar of $N$ is the line through $A$ and $M$.

Next, let $MN$ intersect $PQ$ at $R$. By theorem 5(a) at this link, the points $(N, R, P, Q)$ form a harmonic range. Since the cross ratio of collinear points does not change under central projection,  considering the projection from $A$, $(N,M,B,C)$ also form a harmonic range. (Alternatively, this follows from the theorems of Ceva and Menelaus using the Cevians intersecting at the Gergonne point and transveral $NPQ$). Also, $NE \perp EM$ as both $NI$ and $IE$ are perpendicular to polar $AM$ of $N$.

Considering a central projection from $E$ of line $NMBC$ to a line $N', M, P', C'$ parallel to $NE$ through $M$, we see that $(N', M, P', C')$ form a harmonic range. Since $N'$ is a point at infinity, this implies $M$ is the midpoint of $B'C'$ and so triangles $B'EM$ and $C'EM$ are congruent (equality of two pairs of sides and included angle is $90^{\circ}$). Hence $EM$ bisects $\angle BEC$ as was to be shown.

For the second problem, we use the following characterisation of a symmedian: $AM$ extended concurs with the lines of tangency of the circumcircle at $B$ and $C$. (For three proofs of this see here.)

Define $N$ as the intersection of $XE$ with $BC$ and $D$ as the intersection of $AM$ with the tangents at $B, C$. Note that line $NBMC$ is the polar of $D$ with respect to the circumcircle. By La Hire’s theorem, $D$ must be on the polar of $N$. This polar is perpendicular to $NX$ (the line joining $N$ to the centre of the circle) and as $ED \perp EX$ by construction of $E$, it follows that line $AEMD$ is the polar of $N$. Again by theorem 5(a) in reference (2), $(N, M, B, C)$ form a harmonic range. Following the same argument as the previous proof, this together with $NE \perp EM$ imply $EM$ bisects $\angle BEC$ as required.

By similar arguments, one can prove the following, left to the interested reader. If $X$ is the $A$-excentre of $\triangle ABC$, $M$ the ex-circle’s point of tangency of $BC$, and $E$ the foot of the perpendicular from $X$ to line $AM$, then $EM$ bisects $\angle BEC$.

#### References

(1) Alexander Bogomolny, Poles and Polars from Interactive Mathematics Miscellany and Puzzles http://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtml, Accessed 19 March 2017

(3) Yufei Zhao, Lemmas in Euclidean Geometry

## December 19, 2016

### Some special functions and their applications

Filed under: mathematics — ckrao @ 9:55 am

Here are some notes on special functions and where they may arise. We consider functions in applied mathematics beyond field (four arithmetic operations), composition and inverse operations applied to the power and exponential functions.

#### 1. Bessel and related functions

Bessel functions of the first ($J_{\alpha}(x)$) and second ($Y_{\alpha}(x)$) kind of order $\alpha$ satisfy:

$\displaystyle x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - \alpha^2)y = 0$.

Solutions for integer $\alpha$ arise in solving Laplace’s equation in cylindrical coordinates while solutions for half-integer $\alpha$ arise in solving the Helmholtz equation in spherical coordinates. Hence they come about in wave propagation, heat diffusion and electrostatic potential problems. The functions oscillate roughly periodically with amplitude decaying proportional to $1/\sqrt{x}$. Note that $Y_{\alpha}(x)$ is the second linearly independent solution when $\alpha$ is an integer (for integer $n$, $J_{-n}(x) = (-1)^n J_n(x)$). Also, for integer $n$, $J_n$ has the generating function

$\displaystyle \sum_{n=-\infty}^\infty J_n(x) t^n = e^{(\frac{x}{2})(t-1/t)},$

the integral representations

$\displaystyle J_n(x) = \frac{1}{\pi} \int_0^\pi \cos (n \tau - x \sin(\tau)) \,d\tau = \frac{1}{2 \pi} \int_{-\pi}^\pi e^{i(n \tau - x \sin(\tau))} \,d\tau$

and satisfies the orthogonality relation

$\displaystyle \int_0^1 x J_\alpha(x u_{\alpha,m}) J_\alpha(x u_{\alpha,n}) \,dx = \frac{\delta_{m,n}}{2} [J_{\alpha+1}(u_{\alpha,m})]^2 = \frac{\delta_{m,n}}{2} [J_{\alpha}'(u_{\alpha,m})]^2,$

where $\alpha > -1$, $\delta_{m,n}$ Kronecker delta, and $u_{\alpha, m}$ is the m-th zero of $J_{\alpha}(x)$.

Modified Bessel functions of the first ($I_{\alpha}(x)$) and second ($K_{\alpha}(x)$) kind of order $\alpha$ satisfy:

$\displaystyle x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} - (x^2 + \alpha^2)y = 0$

(replacing $x$ with $ix$ in the previous equation).

The four functions may be expressed as follows.

$\displaystyle J_{\alpha}(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}$

$\displaystyle I_\alpha(x) = \sum_{m=0}^\infty \frac{1}{m! \, \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}$

$\displaystyle Y_\alpha(x) = \frac{J_\alpha(x) \cos(\alpha\pi) - J_{-\alpha}(x)}{\sin(\alpha\pi)}$

$\displaystyle K_\alpha(x) = \frac{\pi}{2} \frac{I_{-\alpha} (x) - I_\alpha (x)}{\sin (\alpha \pi)}$

(In the last formula we need to take a limit when $\alpha$ is an integer.)

Note that $K$ and $Y$ are singular at zero.

The Hankel functions $H_\alpha^{(1)}(x) = J_\alpha(x)+iY_\alpha(x)$ and $H_\alpha^{(2)}(x) = J_\alpha(x)-iY_\alpha(x)$ are also known as Bessel functions of the third kind.

The functions $J_\alpha$$Y_\alpha$, $H_\alpha^{(1)}$, and $H_\alpha^{(2)}$ all satisfy the recurrence relations (using $Z$ in place of any of these four functions)

$\displaystyle \frac{2\alpha}{x} Z_\alpha(x) = Z_{\alpha-1}(x) + Z_{\alpha+1}(x),$
$\displaystyle 2\frac{dZ_\alpha}{dx} = Z_{\alpha-1}(x) - Z_{\alpha+1}(x).$

Bessel functions of higher orders/derivatives can be calculated from lower ones via:

$\displaystyle \left( \frac{1}{x} \frac{d}{dx} \right)^m \left[ x^\alpha Z_{\alpha} (x) \right] = x^{\alpha - m} Z_{\alpha - m} (x),$
$\displaystyle \left( \frac{1}{x} \frac{d}{dx} \right)^m \left[ \frac{Z_\alpha (x)}{x^\alpha} \right] = (-1)^m \frac{Z_{\alpha + m} (x)}{x^{\alpha + m}}.$

In particular, note that $-J_1(x)$ is the derivative of $J_0(x)$.

The Airy functions of the first ($Ai(x)$) and second ($Bi(x)$) kind satisfy

$\displaystyle \frac{d^2y}{dx^2} - xy = 0$.

This arises as a solution to Schrödinger’s equation for a particle in a triangular potential well and also describes interference and refraction patterns.

#### 2. Orthogonal polynomials

Hermite polynomials (the probabilists’ defintion) can be defined by:

$\displaystyle \mathit{He}_n(x)=(-1)^n e^{\frac{x^2}{2}}\frac{d^n}{dx^n}e^{-\frac{x^2}{2}}=\left (x-\frac{d}{dx} \right )^n \cdot 1$,

and are orthogonal with respect to weighting function $w(x) = e^{-x^2}$ on $(-\infty, \infty)$.

They satisfy the differential equation

$\displaystyle \left(e^{-\frac{x^2}{2}}u'\right)' + \lambda e^{-\frac{1}{2}x^2}u = 0$

(where $\lambda$ is forced to be an integer if we insist $u$ be polynomially bounded at $\infty$)

and the recurrence relation

$\displaystyle {\mathit{He}}_{n+1}(x)=x{\mathit{He}}_n(x)-{\mathit{He}}_n'(x)$.

The first few such polynomials are $1, x, x^2-1, x^3-3x, \ldots$. The Physicists’ Hermite polynomials $H_n(x)$ are related by $H_n(x)=2^{\tfrac{n}{2}}{\mathit{He}}_n(\sqrt{2} \,x)$ and arise for example as the eigenstates of the quantum harmonic oscillator.

Laguerre polynomials are defined by

$\displaystyle L_n(x)=\frac{e^x}{n!}\frac{d^n}{dx^n}\left(e^{-x} x^n\right) =\frac{1}{n!} \left( \frac{d}{dx} -1 \right) ^n x^n = \sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} x^k$,

and are orthogonal with respect to $e^{-x}$ on $(0,\infty)$.

They satisfy the differential equation

$\displaystyle xy'' + (1 - x)y' + ny = 0$,

recurrence relation

$\displaystyle L_{k + 1}(x) = \frac{(2k + 1 - x)L_k(x) - k L_{k - 1}(x)}{k + 1}$,

and have generating function

$\displaystyle \sum_n^\infty t^n L_n(x)= \frac{1}{1-t} e^{-\frac{tx}{1-t}}.$

The first few values are $1, 1-x, (x^2-4x+2)/2$. Note also that $L_{-n}(x)=e^xL_{n-1}(-x)$.

The functions come up as the radial part of solution to Schrödinger’s equation for a one-electron atom.

Legendre polynomials can be defined by

$\displaystyle P_n(x) = {1 \over 2^n n!} {d^n \over dx^n } \left[ (x^2 -1)^n \right]$

and are orthogonal with respect to the $L^2$ norm on $(-1,1)$.

They satisfy the differential equation

$\displaystyle {d \over dx} \left[ (1-x^2) {d \over dx} P_n(x) \right] + n(n+1)P_n(x) = 0$,

recurrence relation

and have generating function

$\sum_{n=0}^\infty P_n(x) t^n = \displaystyle \frac{1}{\sqrt{1-2xt+t^2}}$.

The first few values are $1, x, (3x^2-1)/2, (5x^3-3x)/2$.

They arise in the expansion of the Newtonian potential $1/|x-x'|$ (multipole expansions) and Laplace’s equation where there is axial symmetry (spherical harmonics are expressed in terms of these).

Chebyshev polynomials of the 1st kind $T_n(x)$ can be defined by

$T_n(x) =\begin{cases} \cos(n\arccos(x)) & \ |x| \le 1 \\ \frac12 \left[ \left (x-\sqrt{x^2-1} \right )^n + \left (x+\sqrt{x^2-1} \right )^n \right] & \ |x| \ge 1 \\ \end{cases}$

and are orthogonal with respect to weighting function $w(x) = 1/\sqrt{1-x^2}$ in $(-1,1)$.

They satisfy the differential equation

$\displaystyle (1-x^2)\,y'' - x\,y' + n^2\,y = 0$,

the relations

$\displaystyle T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)$

$\displaystyle (1 - x^2)T_n'(x) = -nx T_n(x) + n T_{n-1}(x)$

and have generating function

$\displaystyle \sum_{n=0}^{\infty}T_n(x) t^n = \frac{1-tx}{1-2tx+t^2}.$

The first few values are $1, x, 2x^2-1, 4x^3-3x, \ldots$. These polynomials arise in approximation theory, namely their roots are used as nodes in piecewise polynomial interpolation. The function $f(x) = \frac1{2^{n-1}}T_n(x)$ is the polynomial of leading coefficient 1 and degree n where the maximal absolute value on (-1,1) is minimal.

Chebyshev polynomials of the 2nd kind $U_n(x)$ are defined by

$\displaystyle U_n(x) = \frac{\left (x+\sqrt{x^2-1} \right )^{n+1} - \left (x-\sqrt{x^2-1} \right )^{n+1}}{2\sqrt{x^2-1}}$

and are orthogonal with respect to weighting function $w(x) = \sqrt{1-x^2}$ in $(-1,1)$.

They satisfy the differential equation

$\displaystyle (1-x^2)\,y'' - 3x\,y' + n(n+2)\,y = 0$,

the recurrence relation

$\displaystyle U_{n+1}(x) = 2xU_n(x) - U_{n-1}(x)$

and have generating function

$\displaystyle \sum_{n=0}^{\infty}U_n(x) t^n = \frac{1}{1-2 t x+t^2}.$

The first few values are $1, 2x, 4x^2-1, 8x^3-4x, \ldots$. (There are also less well known Chebyshev  polynomials of the third and fourth kind.)

Bessel polynomials $y_n(x)$ may be defined from Bessel functions via

$\displaystyle y_n(x)=\sqrt{\frac{2}{\pi x}}\,e^{1/x}K_{n+\frac 1 2}(1/x) = \sum_{k=0}^n\frac{(n+k)!}{(n-k)!k!}\,\left(\frac{x}{2}\right)^k$.

They satisfies the differential equation

$\displaystyle x^2\frac{d^2y_n(x)}{dx^2}+2(x\!+\!1)\frac{dy_n(x)}{dx}-n(n+1)y_n(x)=0$.

The first few values are $1, x+1, 3x^2+3x+1,\ldots$.

3. Integrals

The error function has the form

$\displaystyle \rm{erf}(x) = \frac{2}{\sqrt\pi}\int_0^x e^{-t^2}\,\mathrm dt$.

This can be interpreted as the probability a normally distributed random variable with zero mean and variance 1/2 is in the interval $(-x,x)$.

The cdf of the normal distribution $\Phi(x)$ is related to this via $\Phi(x) = (1 + {\rm erf}(x/\sqrt{2})/2$. Hence the tail probability of the standard normal distribution $Q(x)$ is $Q(x) = (1 - {\rm erf}(x/\sqrt{2}))/2$.

Fresnel integrals are defined by

$\displaystyle S(x) =\int_0^x \sin(t^2)\,\mathrm{d}t=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+3}}{(2n+1)!(4n+3)}$
$\displaystyle C(x) =\int_0^x \cos(t^2)\,\mathrm{d}t=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+1}}{(2n)!(4n+1)}$

They have applications in optics.

The exponential integral ${\rm Ei}(x)$ (used in heat transfer applications) is defined by

$\displaystyle {\rm Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}t\,dt$.

It is related to the logarithmic integral

$\displaystyle {\rm li} (x) = \int_0^x \frac{dt}{\ln t}$

by $\mathrm{li}(x) = \mathrm{Ei}(\ln x)$ (for real $x$).

The incomplete elliptic integral of the first, second and third kinds are defined by

$\displaystyle F(\varphi,k) = \int_0^\varphi \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}$

$\displaystyle E(\varphi,k) = \int_0^\varphi \sqrt{1-k^2 \sin^2\theta}\, d\theta$

$\displaystyle \Pi(n ; \varphi \setminus \alpha) = \int_0^\varphi \frac{1}{1-n\sin^2 \theta} \frac {d\theta}{\sqrt{1-(\sin\theta\sin \alpha)^2}}$

Setting $\varphi = \pi/2$ gives the complete elliptic integrals.

Any integral of the form $\int_{c}^{x} R \left(t, \sqrt{P(t)} \right) \, dt$, where $c$ is a constant, $R$ is a rational function of its arguments and $P(t)$ is a polynomial of 3rd or 4th degree with no repeated roots, may be expressed in terms of the elliptic integrals. The circumference of an ellipse of semi-major axis $a$, semi-minor axis $b$ and eccentricity $e = \sqrt{1-b^2/a^2}$ is given by $4aE(e)$, where $E(k)$ is the complete integral of the second kind.

(Some elliptic functions are related to inverse elliptic integral, hence their name.)

The (upper) incomplete Gamma function is defined by

$\displaystyle \Gamma(s,x) = \int_x^{\infty} t^{s-1}\,e^{-t}\,{\rm d}t$.

It satisfies the recurrence relation $\Gamma(s+1,x)= s\Gamma(s,x) + x^{s} e^{-x}$. Setting $s= 0$ gives the Gamma function which interpolates the factorial function.

The digamma function is the logarithmic derivative of the gamma function:

$\displaystyle \psi(x)=\frac{d}{dx}\ln\Big(\Gamma(x)\Big)=\frac{\Gamma'(x)}{\Gamma(x)}$.

Due the relation $\psi(x+1) = \psi(x) + 1/x$, this function appears in the regularisation of divergent integrals, e.g.

$\sum_{n=0}^{\infty} \frac{1}{n+a}= - \psi (a)$.

The incomplete Beta function is defined by

$\displaystyle B(x;\,a,b) = \int_0^x t^{a-1}\,(1-t)^{b-1}\,\mathrm{d}t$.

When setting $x=1$ this becomes the Beta function which is related to the gamma function via

$\displaystyle B(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$.

This can be extended to the multivariate Beta function, used in defining the Dirichlet function.

$\displaystyle B(\alpha_1,\ldots,\alpha_K) = \frac{\Gamma(\alpha_1) \cdots \Gamma(\alpha_K)}{\Gamma(\alpha_1 + \ldots + \alpha_K)}$.

The polylogarithm, appearing as integrals of the Fermi–Dirac and Bose–Einstein distributions, is defined by

$\displaystyle {\rm Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s} = z + \frac{z^2}{2^s} + \frac{z^3}{3^s} + \cdots$

Note the special case ${\rm Li}_1(z) = -\ln (1-z)$ and the case $s=2$ is known as the dilogarithm. We also have the recursive formula

$\displaystyle {\rm Li}_{s+1}(z) = \int_0^z \frac {{\rm Li}_s(t)}{t}\,\mathrm{d}t$.

#### 4. Generalised Hypergeometric functions

All the above functions can be written in terms of generalised hypergeometric functions.

$\displaystyle {}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z) = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}$

where $(a)_n = \Gamma(a+n)/\Gamma(a) = a(a+1)(a+2)...(a+n-1)$ for $n > 0$ or $(a)_0 = 1$.

The special case $p=q=1$ is called a confluent hypergeometric function of the first kind, also written $M(a;b;z)$.

This satisfies the differential equation (Kummer’s equation)

$\displaystyle \left (z\frac{d}{dz}+a \right )w = \left (z\frac{d}{dz}+b \right )\frac{dw}{dz}$.

The Bessel, Hankel, Airy, Laguerre, error, exponential and logarithmic integral functions can be expressed in terms of this.

The case $p=2, q=1$ is sometimes called Gauss’s hypergeometric functions, or simply hypergeometric functions. This satisfies the differential equation

$\displaystyle \left (z\frac{d}{dz}+a \right ) \left (z\frac{d}{dz}+b \right )w =\left (z\frac{d}{dz}+c \right )\frac{dw}{dz}$.

The Legendre, Hermite and Chebyshev, Beta, Gamma functions can be expressed in terms of this.

The Wolfram Functions Site

Wikipedia: List of mathematical functions

Wikipedia: List of special functions and eponyms

Wikipedia: List of q-analogs

Wikipedia Category: Orthogonal polynomials

Weisstein, Eric W. “Laplace’s Equation.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/LaplacesEquation.html

## June 26, 2016

### 2016 has many factors

Filed under: mathematics — ckrao @ 11:16 am

The number 2016 has at least as many factors (36) as any positive integer below it except $1680 = 2^4\times 3\times 5\times 7$ (which has 40 factors). The next time a year will have more factors is $2160 = 2^4\times 3^3\times 5$, also with 40 factors.

Here are the numbers below 2160 also with 36 factors:

• $1260 = 2^2 \times 3^2 \times 5 \times 7$
• $1440 = 2^5 \times 3^2 \times 5$
• $1800 = 2^3 \times 3^2 \times 5^2$
• $1980 = 2^2 \times 3^2 \times 5 \times 11$
• $2016 = 2^5 \times 3^2 \times 7$
• $2100 = 2^2 \times 3 \times 5^2 \times 7$

The first integer with more than 40 factors is $2520 = 2^3 \times 3^2 \times 5 \times 7$ (48 factors).

#### References

[1] N. J. A. Sloane and Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 10000 terms from N. J. A. Sloane), A000005 – OEIS.

## March 26, 2016

### Applying AM-GM in the denominator after flipping the sign

Filed under: mathematics — ckrao @ 8:44 pm

There are times when solving inequalities that one has a sum of fractions in which applying the AM-GM inequality to each denominator results in the wrong sign for the resulting expression.

For example (from [1], p18), if we wish to show that for real numbers $x_1, x_2, \ldots, x_n$ with sum $n$ that

$\displaystyle \sum_{i = 1}^n \frac{1}{x_i^2 + 1}\geq \frac{n}{2},$

we may write $x_i^2 + 1 \geq 2x_i$ (equivalent to $(x_i-1)^2 \geq 0$), but this implies $\frac{1}{x_i^2 + 1} \leq \frac{1}{2x_i}$ and so the sign goes the wrong way.

A way around this is to write

\begin{aligned} \frac{1}{x_i^2 + 1} &= 1 - \frac{x_i^2}{x_i^2 + 1}\\ &\geq 1 - \frac{x_i^2}{2x_i}\\ &= 1 - \frac{x_i}{2}. \end{aligned}

Summing this over $i$ then gives $\sum_{i=1}^n \frac{1}{x_i^2 + 1} \geq n - \sum_{i=1}^n (x_i/2) = n/2$ as desired.

Here are a few more examples demonstrating this technique.

2. (p9 of [2]) If $a,b,c$ are positive real numbers with $a + b + c = 3$, then

$\dfrac{a}{1 +b^2} + \dfrac{b}{1 +c^2} + \dfrac{c}{1 +a^2} \geq \dfrac{3}{2}.$

To prove this we write

\begin{aligned} \frac{a}{1 + b^2} &= a\left(1 - \frac{b^2}{1 + b^2}\right)\\ &\geq a\left(1 - \frac{b}{2}\right) \quad \text{(using the same argument as before)}\\ &=a - \frac{ab}{2}. \end{aligned}

Next we have $3(ab + bc + ca) \leq (a + b + c)^2 = 9$ as this is equivalent to $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$. This means $ab + bc + ca \leq 3$. Putting everything together,

\begin{aligned} \frac{a}{1 + b^2} + \frac{b}{1 + c^2} + \frac{c}{1 + a^2}&\geq \left( a - \frac{ab}{2} \right) + \left( b - \frac{bc}{2} \right) + \left( c - \frac{ca}{2} \right)\\ &= (a + b + c) - (ab + bc + ca)/2\\ &\geq 3 - 3/2\\ &=\frac{3}{2}, \end{aligned}

as required.

3. (based on p8 of [2]) If $x_i > 0$ for $i= 1, 2, \ldots, n$ and $\sum_{i = 1}^n x_i^2 = n$ then

$\displaystyle \sum_{i=1}^n \frac{1}{x_i^3 + 2} \geq \frac{n}{3}.$

By the AM-GM inequality, $x_i^3 + 2 = x_i^3 + 1 + 1 \geq 3x_i$, so

\begin{aligned} \frac{1}{x_i^3 + 2} &= \frac{1}{2}\left( 1 - \frac{x_i^3}{x_i^3 + 2} \right)\\ &\geq \frac{1}{2}\left( 1 - \frac{x_i^3}{3x_i} \right)\\ &= \frac{1}{2}\left( 1 - \frac{x_i^2}{3} \right). \end{aligned}

Summing this over $i$ gives

\begin{aligned} \sum_{i=1}^n \frac{1}{x_i^3 + 2} &\geq \frac{1}{2} \sum_{i=1}^n \left( 1 - \frac{x_i^2}{3} \right)\\ &= \frac{1}{2}\left( n - \frac{n}{3} \right)\\ &= \frac{n}{3}. \end{aligned}

4. (from [3]) If $x, y, z$ are positive, then

$\dfrac {x ^ 3}{x ^ 2 + y ^ 2} + \dfrac {y ^ 3}{y ^ 2 + z ^ 2} + \dfrac {z ^ 3}{z ^ 2 + x ^ 2 } \geq \dfrac {x + y + z}{2}.$

Once again, focusing on the denominator,

\begin{aligned} \dfrac {x ^ 3}{x ^ 2 + y ^ 2} &= x\left(1 - \dfrac {y ^ 2} {x ^ 2 + y ^ 2} \right)\\ &\geq x \left(1 -\dfrac{xy^2}{2xy} \right)\\ &= x-\dfrac{y}{2}. \end{aligned}

Hence,

\begin{aligned} \dfrac {x ^ 3}{x ^ 2 + y ^ 2} + \dfrac {y ^ 3}{y ^ 2 + z ^ 2} + \dfrac {z ^ 3}{z ^ 2 + x ^ 2 } &\geq x-\dfrac{y}{2} + y-\dfrac{z}{2} + z-\dfrac{x}{2}\\ &= \dfrac {x + y + z}{2}, \end{aligned}

as desired.

5. (from the 1991 Asian Pacific Maths Olympiad, see [4] for other solutions) Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be positive numbers with $\sum_{i = 1}^n a_i = \sum_{i = 1}^n b_i$. Then

$\displaystyle\sum_{i=1}^n\frac{a_i^2}{a_i + b_i} \geq \frac{1}{2}\sum_{i=1}^n a_i.$

Here we write

\begin{aligned} \sum_{i=1}^n\frac{a_i^2}{a_i + b_i} &= \sum_{i=1}^n a_i \left(1 - \frac{b_i}{a_i + b_i} \right)\\ &\geq \sum_{i=1}^n a_i \left(1 - \frac{b_i}{2\sqrt{a_i b_i}} \right) \\ &= \frac{1}{2} \sum_{i=1}^n \left( 2a_i - \sqrt{a_i b_i} \right) \\ &= \frac{1}{4} \sum_{i=1}^n \left( 4a_i - 2\sqrt{a_i b_i} \right)\\ &= \frac{1}{4} \sum_{i=1}^n \left( 2a_i +a_i - 2\sqrt{a_i b_i} + b_i \right) \quad \text{(as } \sum_{i=1}^n a_i = \sum_{i=1}^n b_i\text{)}\\ &= \frac{1}{4} \sum_{i=1}^n \left( 2a_i + \left(\sqrt{a_i} - \sqrt{b_i}\right)^2 \right)\\ &\geq \frac{1}{4} \sum_{i=1}^n 2a_i\\ &= \frac{1}{2} \sum_{i=1}^n a_i, \end{aligned}

as required.

#### References

[1] Zdravko Cvetkovski, Inequalities: Theorems, Techniques and Selected Problems, Springer, 2012.

## February 27, 2016

### Cutting a triangle in half

Filed under: mathematics — ckrao @ 9:40 pm

Here is a cute triangle result that I’m surprised I had not known previously. If we are given a point on one of the sides of a triangle, how do we find a line through the triangle that cuts its area in half?

Clearly if that point is either a midpoint or one of the vertices, the answer is a median of the triangle. A median cuts a triangle in half since the two pieces have the same length side and equal height.

So what if the point is not a midpoint or a vertex? Referring to the diagram below, if $P$ is our desired point closer to $A$ than $B$, the end point $Q$ of the area-bisecting segment would need to be on side $BC$ so that area(BPQ) = area(ABC)/2.

In other words, we require area(BPQ) = area(BDQ), or, subtracting the areas of triangle BDQ from both sides,

$\displaystyle |DPQ| = |DCQ|.$

Since these two triangles share the common base $DQ$, this tells us that we require them to have the same height. In other words, we require $CP$ to be parallel to $DQ$. This tells us how to construct the point $Q$ given $P$ on $AB$:

1. Construct the midpoint D of $AB$.
2. Draw $DQ$ parallel to $AP$.

See [1] for an animation of this construction.

In turns out that the set of all area-bisecting lines are tangent to three hyperbolas and enclose a deltoid of area $(3/4)\ln(2) - 1/2 \approx 0.01986$ times the original triangle. [2,3,4]

#### References

[1] Jaime Rangel-Mondragon, “Bisecting a Triangle” http://demonstrations.wolfram.com/BisectingATriangle/ from the Wolfram Demonstrations Project Published: July 10, 2013

[4] Henry Bottomley, Area bisectors of a triangle, January 2002

## January 30, 2016

### Patterns early in the digits of pi

Filed under: mathematics — ckrao @ 8:59 pm

Recently when taking a look at the early decimal digits of $\pi$ I made the following observations:

3.141592653589793238462643383279502884197169399375105820974944592307816406286…

• The first run of seven distinct digits (8327950, shown underlined) appears in the 26th to 32nd decimal place. Curiously the third such run (5923078, also underlined) in decimal places 61 to 67 contains the same seven digits. (There is also a run of seven distinct digits in places 51 to 57 with 5820974.)
• Decimal digits 60 to 69 (shown in bold) are distinct (i.e. all digits are represented once in this streak). The same is true for digits 61 to 70 as both digits 60 and 70 are ‘4’.

Assume the digits of $\pi$ are generated independently from a uniform distribution. Firstly, how often would we expect to see a run of 7 distinct digits? Places $k$ to $k+6$ are distinct with probability

$\displaystyle \frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} \times \frac{6}{10} \times \frac{5}{10} \times \frac{4}{10} = \frac{9!}{3.10^6} = \frac{189}{3125}.$

Hence we expect runs of 7 distinct digits to appear $3125/189 \approx 16.5$ places apart. This includes the possibility of runs such as 12345678 which contain two runs of 7 distinct digits that are only 1 place apart.

How often would we expect to see the same 7 digits appearing in a run as we did in places 26-32 and 61-67? Furthermore let’s assume the two runs have no overlap, so we discount possibilities such as 12345678 which have a six-digit overlap. We expect a given sequence (e.g. 1234567, in that order) to appear $1/10^7$ of the time. There are $7! = 5040$ permutations of such a sequence, but of these $1$ has overlap 6 (2345671), $2!$ has overlap 5 (3456712 or 3456721), $3!$ has overlap 4, …, $6!$ has overlap 1 with the original sequence. This leaves us with $5040 - (1 + 2 + 6 + 24 + 120 + 720) = 4167$ possible choices of the next run to have the same 7 digits but non-overlapping (or to appear in precisely the same order – overlap 7). Hence we expect the same 7 digits to recur (no overlap with the original run) after approximately $10^7/4167 \approx 2400$ places apart so what we saw in the first 100 places was remarkable.

Now let us turn to runs of all ten distinct digits. Repeating the argument above, such runs occur every $10^{10}/10! \approx 2756$ places. According to [1] the next time we see ten distinct digits is in decimal places 5470 to 5479.

To answer the question of when we would expect to see the first occurrence of ten distinct digits, we adopt an argument from renewal-reward theory based on Sec 7.9.2 of [2] (there also exist approaches based on setting up recurrence relations, or martingale theory (modelling a fair casino), see [3]-[4]). Firstly we let $T$ be the first time we get 10 consecutive distinct values – we wish to find $E[T]$ where $E$ denotes the expected value operator. Note that this will be more than the 2756 answer we obtained above since we make no assumption of starting with a run of ten distinct digits – there is no way $T$ could be 1 for example, but we could have two runs of ten distinct digits that are 1 apart.

From a sequence of digits we first define a renewal process in which after we get 10 consecutive distinct values (at time $T$) we start over and wait for the next run of 10 consecutive distinct values without using any of the values  up to time $T$. Such a process will then have an expected length of cycle of $E[T]$.

Next, suppose we earn a reward of $1 every time the last 10 digits of the sequence are distinct (so we would have obtained$1 at each of decimal places 69 and 70 in the $\pi$ example). By an important result in renewal-reward theory, the long run average reward is equal to the expected reward in a cycle divided by the expected length of a cycle.

In a cycle we will obtain

• $1 at the end •$1 at time 1 in the cycle with probability $1/10$ (if that digit is the same as ten digits before it)
• $1 at time 2 in the cycle with probability $2/100$ (if the last two digits match those ten places before it) •$1 at time 9 in the cycle with probability $9!/10^9$ (if the last nine digits match those ten places before it)

Hence the expected reward in a cycle is given by

$\displaystyle 1 + \sum_{i=1}^9 \frac{i!}{10^i} = \sum_{i=0}^9 \frac{i!}{10^i}.$

We have already seen that the long run average reward is $10!/10^{10}$ at each decimal place. Hence the expected length of a cycle $E[T]$ (i.e. the expected number of digits before we expect the first run of ten consecutive digits) is given by

$\frac{10^{10}}{10!}\sum_{i=0}^9 \frac{i!}{10^i} \approx 3118.$

Hence it is pretty cool that we see it so early in the decimal digits of $\pi$. 🙂

#### References

[2] S. Ross, Introduction to Probability Models, Academic Press, 2014.

[4] A Collection of Dice Problemsmadandmoonly.com

## December 23, 2015

### Areas of sections of a triangle from distances to its sides

Filed under: mathematics — ckrao @ 12:35 pm

If a point $P$ is in the interior of triangle $ABC$ distance $x, y$ and $z$ from the sides, what is the ratio of the area of quadrilateral $BXPZ$ to that of $ABC$?

One way of determining this is to draw parallels to the sides of the triangles through $P$. Let $X_1$ and $X_2$ be where these parallels meet side $BC$ as shown below.

Let the sides of the triangles have lengths $a, b, c$ with corresponding altitudes $h_a, h_b, h_c$.

Then as $\triangle PX_1 X_2$ and $\triangle ACB$ are similar,

\begin{aligned}|PX_1X| &= |PX_1X_2| \frac{X_1X}{X_1X_2}\\ &= |PX_1X_2| \frac{b\cos C}{a}\\ &= |PX_1X_2| \frac{b (a^2 + b^2 - c^2)}{2a^2b} \quad \text{ (cosine rule)}\\ &= \left(\frac{x}{h_a} \right)^2|ABC|\frac{(a^2 + b^2 - c^2)}{2a^2}\\ &= |ABC|\left(\frac{ax}{ax+by+cz}\right)^2\frac{(a^2 + b^2 - c^2)}{2a^2}\\&= \frac{|ABC|x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2},\quad\quad (1) \end{aligned}

where the second last line follows from twice the area of |ABC| being $ah_a = ax + by + cz$.

Similarly,

$\displaystyle |PY_1Z| = \frac{|ABC|z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}.\quad \quad (2)$

Finally,

\begin{aligned}|X_1Y_1B| &= \left(\frac{h_b-y}{h_b}\right)^2|ABC|\\ &= \left(1-\frac{by}{bh_b}\right)^2 |ABC|\\ &= \left(1-\frac{by}{2|ABC|}\right)^2|ABC|\\ &= \left(1-\frac{by}{ax+by+cz}\right)^2|ABC|\\ &= \left(\frac{ax +cz}{ax+by+cz}\right)^2|ABC|. \quad\quad(3)\end{aligned}

Combining (1), (2) and (3), we obtain our desired answer as

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{|X_1Y_1B|-|PX_1X|-|PY_1Z|}{|ABC|}\\&= \left(\frac{ax +cz}{ax+by+cz}\right)^2-\frac{x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2}-\frac{z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}\\&=\frac{(ax+cz)^2 - x^2(a^2 +b^2-c^2)/2 - z^2(b^2+c^2-a^2)/2}{(ax+by+cz)^2}\\ &= \frac{2axcz + x^2(a^2 - b^2 + c^2) + z^2(c^2 +a^2-b^2)}{(ax+by+cz)^2}\\&= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}.\quad\quad(4)\end{aligned}

Similar formulas can be found for quadrilaterals $XPYC$ and $YPZA$ by permuting variables. Note that if $P$ is outside the triangle or if the triangle is obtuse-angled, care must be taken in the signs of the areas (the quadrilaterals may not be convex) and variables $x, y, z$.

Note that (4) may also be written as

$\displaystyle \frac{|BXPZ|}{|ABC|} = \frac{ac(2xz + (x^2 + z^2)\cos B)}{(ax+by+cz)^2}.\quad\quad(5)$

### Special cases

1) If $\triangle ABC$ is equilateral, $a=b=c$ and from (4) we obtain

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(6)\end{aligned}

2) If $P$ is at the incentre of $\triangle ABC$, then $x = y = z = r$ (the inradius) and from (4) we have

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(7)\end{aligned}

3) If $\triangle P$ is right-angled at $B$, then quadrilateral $BXPZ$ is a rectangle with area $xz$ and $\triangle ABC$ has area $ac/2$ and from (5),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{2acxz )}{(ax+by+cz)^2}\\ &= \frac{2acxz )}{(ac)^2}\\ &= \frac{2xz}{ac}.\quad \quad (8)\end{aligned}

as expected.

4) If $a=c$ and $x=z$ (symmetric isosceles triangle case) then from (4),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2x^2 + 2x^2(2a^2-b^2)}{2(2ax+by)^2}\\ &= \frac{x^2(4a^2 -b^2)}{(2ax+by)^2}.\quad\quad(9)\end{aligned}

## November 25, 2015

### An identity based on three numbers summing to zero

Filed under: mathematics — ckrao @ 11:06 am

Here is a nice identity which according to [1] appeared in a 1957 Chinese mathematics competition.

If $x + y + z = 0$ then

$\displaystyle \left(\frac{x^2 + y^2 + z^2}{2} \right)\left(\frac{x^5 + y^5 + z^5}{5} \right) = \left(\frac{x^7 + y^7 + z^7}{7} \right).\quad \quad (1)$

An elegant proof of this avoids any lengthy expansions. Let $x$, $y$ and $z$ be roots of the cubic polynomial

\begin{aligned} (X-x)(X-y)(X-z) &= X^3 - (x+y+z)X^2 + (xy + yz + zx)X - xyz\\ &:= X^3 + aX + b.\quad \quad (2)\end{aligned}

Then

\begin{aligned} x^2 + y^2 + z^2 &= (x+y+z)^2 - 2(xy + yz + xz)\\ &= 0 -2a\\ &= -2a\quad\quad (3)\end{aligned}

and summing the relation $X^3 = -aX - b$ for each of $X=x, X=y$ and $X=z$, gives

\begin{aligned}x^3 + y^3 + z^3 &= -a(x + y + z) - 3b\\ &= -3b.\quad\quad (4)\end{aligned}

In a similar manner, $X^4 = -aX^2 - bX$ and so

\begin{aligned} x^4 + y^4 + z^4 &= -a(x^2 + y^2 + z^2) - b(x + y + z)\\ &= -a(-2a)\\ &= 2a^2.\quad \quad (5)\end{aligned}

Next, $X^5 = -aX^3 - bX^2$ and so

\begin{aligned} x^5 + y^5 + z^5 &= -a(x^3 + y^3 + z^3) - b(x^2 + y^2 + z^2)\\ &= -a(-3b) -b(-2a)\\ &= 5ab.\quad \quad (6)\end{aligned}

Finally, $X^7 = -aX^5 - bX^4$ and so

\begin{aligned} x^7 + y^7 + z^7 &= -a(x^5 + y^5 + z^5) - b(x^4 + y^4 + z^4)\\ &= -a(5ab) -b(2a^2)\\ &= -7a^2b.\quad \quad (7)\end{aligned}

We then combine (3), (6) and (7) to obtain (1). It seems that $x^n + y^n + z^n$ for higher values of $n$ are more complicated expressions in $a$ and latex $b$, so we don’t get as pretty a relation elsewhere.

#### Reference

[1] Răzvan Gelca and Titu Andreescu, Putnam and Beyond, Springer, 2007.

## October 28, 2015

### The product of distances to a point from vertices of a regular polygon

Filed under: mathematics — ckrao @ 11:12 am

Here is a cool trigonometric identity I recently encountered:

$\displaystyle \prod_{k=1}^n \sin \frac{(2k-1)\pi}{4n} = \prod_{k=1}^n \cos \frac{(2k-1)\pi}{4n} = \frac{\sqrt{2}}{2^n}.$

For example, for $n = 9$:

$\displaystyle \sin 5^{\circ} \sin 15^{\circ} \sin 25^{\circ} \ldots \sin 85^{\circ} = \frac{\sqrt{2}}{2^9}.$

After thinking about it for some time I realised that the terms on the left side can each be seen as half the lengths of chords of a unit circle with 4n evenly spaced points that can then be rearranged to be distances from a point on the unit circle to half of the points of a regular (2n)-gon, as shown in the figure below.

With the insight of this figure we then write

\begin{aligned} \prod_{k=1}^n \sin \frac{(2k-1)\pi}{4n} &= \left(\prod_{k=1}^{2n} \left| \sin \frac{(2k-1)\pi}{4n} \right|\right)^{1/2} \quad \text{ (all terms are positive)}\\ &= \prod_{k=1}^{2n} \left| \frac{\exp(i\frac{(2k-1)\pi}{4n}) - \exp(-i\frac{(2k-1)\pi}{4n})}{2i} \right|^{1/2} \\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \exp\left(-i \frac{(2k+1)\pi}{4n}\right) \left(\exp\left(i\frac{4k\pi}{4n}\right) - \exp\left(i\frac{2\pi}{4n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \exp\left(-i \frac{(2k+1)\pi}{4n}\right) \right|^{1/2} \left| \left(\exp\left(i\frac{4k\pi}{4n}\right) - \exp\left(i\frac{2\pi}{4n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \left(\exp\left(i\frac{2k\pi}{2n}\right) - \exp\left(i\frac{\pi}{2n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \left|\prod_{k=1}^{2n} \left(z-\exp\left(i\frac{2k\pi}{2n}\right)\right) \right|^{1/2} \quad \text{where }z = \exp\left(i\frac{\pi}{2n}\right)\\ &= \frac{1}{2^n} |(z^{2n}-1)|^{1/2}\\ &= \frac{1}{2^n} |-1-1|^{1/2}\\ &= \frac{\sqrt{2}}{2^n}. \end{aligned}

(The cosine formula can be derived in a similar manner.)

In general, the product of the distances of any point $z$ in the complex plane to the n roots of unity $\omega_n$  is

$\displaystyle \prod_{k=0}^{n-1} |z-\omega_n| = |z^n - 1|.$

The above case was where $z^n = -1$. Two more cases are illustrated below, this time for n = 10. In the left example the product of distances is

$\displaystyle \prod_{k=0}^9 |(1+i)-\exp(2\pi i k/10)| = |(1+i)^{10}-1| = 5\sqrt{41}$

while for the right example it is

$\displaystyle \prod_{k=0}^9 |1/2-\exp(2\pi i k/10)| = |(1/2)^{10}-1| = 1023/1024.$

Note that earlier in the year I posted on the distances to a line from vertices of a regular polygon.

## September 20, 2015

### The simplest Heronian triangles

Filed under: mathematics — ckrao @ 12:05 pm

Heronian triangles are those whose side lengths and area have integer value. Most of the basic ones are formed either by right-angled triangles of integer sides, or by two such triangles joined together. Following the proof in [1] it is not difficult to show that such triangles have side lengths proportional to $(x,y,z) = (n(m^2 + h^2), m(n^2 + h^2), (m+n)(mn-h^2))$ where $m,n$ and $h$ are integers with $mn > h^2$.  Firstly, if a triangle has integer side lengths and area, its altitudes must be rational, being twice the area divided by a side length. Also by the cosine rule, the cosine of its angles must be rational, so $z_1$ and $z_2$ in the diagram below are rational too (here assume $z$ is the longest side, so that the altitude is inside the triangle).

This gives us the equations

$\displaystyle h^2 = x^2 - z_1^2 = y^2 - z_2^2, z_1 + z_2 = z,\quad \quad (1)$

where $h, z_1, z_2 \in \mathbb{Q}$. Letting $x + z_1 = m$ and $y + z_2 = n$ it follows from the above equations that $x - z_1 = h^2/m, y-z^2 = h^2/n$ from which

$\displaystyle (x,y,z) = \left(\left(\frac{1}{2}(m + \frac{h^2}{m}\right), \frac{1}{2}\left(n + \frac{h^2}{n}\right), \frac{1}{2}\left( m - \frac{h^2}{m} + (n - \frac{h^2}{n}\right)\right). \quad\quad (2)$

Scaling the sides up by a factor of $2mn$, the sides are proportional to

$(x',y',z') = (n(m^2 + h^2), m(n^2 + h^2), (m+n)(mn-h^2)).\quad\quad(3)$

Next, letting $d$ be the common denominator of the rational numbers $h, z_1$ and $z_2$, we multiply the rational solution $(x', y', z')$ in (3) each by $d^3$ to obtain an integral solution. The altitude upon side length $z$ is proportional to $2hmn$ and the area is $hmn(m+n)(mn-h^2)$. Hence if we start with positive $m,n,h$ with no common factor and with $mn > h^2$, then (3) gives the side lengths of a Heronian triangle that can then be made primitive by dividing by a common factor.

Below the 20 primitive Heronian triangles with area less than 100 are illustrated to scale, where the first row has been doubled in size for easier viewing (a larger list is here). Note that all but one of them is either an integer right-angled triangle or decomposable into two such triangles as indicated by the blue numbers and sides. Refer to [2] for more on triangles which are not decomposable into two integer right-angled triangles. Here are the primitive Pythagorean triples that feature in the triangles:

• 3-4-5
• 5-12-13
• 8-15-17
• 20-21-29
• 7-24-25
• 28-45-53

#### References

[1] Carmichael, R. D., 1914, “Diophantine Analysis”, pp.11-13; in R. D. Carmichael, 1959, The Theory of Numbers and Diophantine Analysis, Dover Publications, Inc.

[2] Yiu, Paul (2008), Heron triangles which cannot be decomposed into two integer right triangles (PDF), 41st Meeting of Florida Section of Mathematical Association of America.

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