# Math Challenge - May 2021

• Challenge
Mentor
Summary: Group Theory, Integrals, Representation Theory, Iterations, Geometry, Abstract Algebra, Linear Algebra.

1. Integrate
$$\int_{0}^\infty \int_{0}^\infty e^{-\left(x+y+\frac{\lambda^3 }{xy}\right)} x^{-\frac{2}{3}}y^{-\frac{1}{3}}\,dx\,dy$$

2. (solved by @Infrared , basic solution still possible ) Let ##F_n## be the free group of rank ##n## with generators ##\{w_1,\ldots,w_n\}.## Then
$$\prod_{i=1}^m w_{a_i}^{b_i} \in [F_n,F_n] \Longleftrightarrow \forall_{k=1}^m \;\sum_{a_i=k}b_i =0$$

3. Calculate
$$\int_0^\pi \int_0^\pi \int_0^\pi \dfrac{1}{1-\cos x\,\cos y\,\cos z}\,dx\,dy\,dz$$

4. (solved by @fishturtle1 ) Let ##G## be a finite group, ##\mathbb{K}## a field such that ##\operatorname{char}(\mathbb{K})\nmid |G|,## and ##(\rho,V)## and ##(\tau,W)## linear representations of ##G## over ##\mathbb{K}.## The ##\mathbb{K}-##linear mapping
\begin{align*}
\operatorname{Sym}\, : \,\operatorname{Hom}_\mathbb{K}(V,W)&\longrightarrow
\operatorname{Hom}_\mathbb{K}(V,W)\\
\varphi &\longmapsto \operatorname{Sym}(\varphi)=\dfrac{1}{|G|}\sum_{g\in G}\tau(g)\circ\varphi \circ\rho(g^{-1})
\end{align*}
is a projection onto the subspace
$$\operatorname{Hom}_\mathbb{K}((\rho,V),(\tau,W))=\{\vartheta :V\longrightarrow W\,|\,\forall_{g\in G}\, : \,\tau(g)\circ\vartheta\circ \rho(g^{-1})=\vartheta \}$$
of ## \operatorname{Hom}_\mathbb{K}(V,W).## Determine and prove (mention) all five claims.

5. Let ##f(x)=x^3-\dfrac{49}{6}x^2+\dfrac{39}{2}x-\dfrac{31}{3}##. Prove that there are at least one ##a,b## such that ##f^2(a)=a\, , \,f(a)\neq a ## and ##f^4(b)=b\, , \,f^k(b)\neq b\;(k<4)## where ##f^n:=f\circ f^{n-1}, f^1=f.##
Is this true for every even power?

6. (solved by @etotheipi ) Prove the equivalence of the theorem of Pythagoras with the following transversal theorem about isosceles triangles:

Given an isosceles triangle ##\triangle ABC## with baseline ##\overline{AB}\subseteq g,## peak ##C,## i.e. ##|AC|=|BC|,## and ##g## the straight along the baseline. Moreover let ##P\in g## be an arbitrary point. Then
\begin{align*}
|CP|^{2}&=|CA|^{2}+|PA|\cdot |PB|\text{ if }P\not \in \overline{AB}\\
|CP|^{2}&=|CA|^{2}-|PA|\cdot |PB|\text{ if }P \in \overline{AB}
\end{align*}

7. Let ##\alpha ## be an algebraic number of degree ##n\geq 1.## Then there is a real number ##c>0## such that for all ##\mathbb{Q}\ni\dfrac{p}{q}\neq \alpha ##
$$\left|\alpha -\dfrac{p}{q}\right|\geq \dfrac{c}{q^n}$$

8. (solved by @etotheipi ) Let ##a_{n+1}=2+\sqrt{4+a_n}\, , \,a_0\geq -4\,,## be a sequence of real numbers. Determine - if existent - its limit in dependence of the initial value ##a_0,## and show that ##a_n\in [2,5]## in cases where ##a_0\in[-4,5],## and ##a_n\geq 5## in cases where ##a_0\geq 5## ##(n\in \mathbb{N}).##

9. (solved by @etotheipi ) Calculate center, foci, semi-axis, and area of the maximal inscribed ellipse of the triangle ##(1,1),(5,2),(3,6).##

10. (solved by @julian ) Let ##A,B\in \mathbb{M}(n,\mathbb{F})## be two square ##n\times n ## matrices over a field ##\mathbb{F}.## Show that the minimal polynomials of ##AB## and ##BA## are the same in case ##A## is regular. Is it true as well, if ##A## is singular?

High Schoolers only (until 26th)

11.
(solved by @kshitij ) For which positive real numbers ##\mathbb{R}\ni a,b>0 ## does
$$f(a,b)=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}+\dfrac{a}{b}+\dfrac{b}{a}$$
assume a minimal value, and which one?

12. (solved by @kshitij ) Find all pairs ##(x,y)## of integers such that
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$

13. Show that
$$\underbrace{\left|x-\dfrac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\right|}_{=:f(x)}\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$$
You may use ##\pi =3.14159+\delta \, , \,\sqrt{2}=1.41421+\varepsilon ## with ##\delta,\varepsilon \in \left(0,10^{-5}\right).##

14. If ##f(x)=a_nx^n+\ldots+a_1x+a_0\in \mathbb{R}[x]## is a real polynomial of degree ##n## which doesn't have real zeros, and ##h\in\mathbb{R}## a real number, then
$$F(x):=f(x)+h\cdot f'(x)+h^2\cdot f''(x)+\ldots+h^n\cdot f^{(n)}(x)$$
doesn't have real zeros either.

15. (solved by @kshitij ) Solve the following real equations system:
\begin{align*}
x+y&=az\\
x-y&=bz\\
x^2+y^2&=cz
\end{align*}

Last edited:
kshitij and etotheipi

etotheipi
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Problem 6. Let the angle ##\angle CAB = \angle CBA := \theta##. First consider ##P \in \overline{AB}##, then by the cosine rule $$|CP|^2 = |CA|^2 + |AP|^2 - 2|CA|\cdot|AP| \cos{\theta}$$Since ##|AB| = 2|CA| \cos{\theta}## then $$|AP|^2 - 2|CA| \cdot|AP| \cos{\theta} = |AP| \cdot\left( |AP| - 2 |CA| \cos{\theta} \right) = |AP|\cdot \left( |AP| - |AB| \right) = -|AP|\cdot |BP|$$hence ##|CP|^2 = |CA|^2 - |AP|\cdot |BP|##.

It's similar for ##P \not \in \overline{AB}##. WLOG take ##P## to be on the left of ##A##. Then\begin{align*}

|CP|^2 &= |CA|^2 + |AP|^2 - 2|CA|\cdot |AP| \cos{(\pi - \theta)} \\

&= |CA|^2 + |AP|\cdot \left( |AP| + 2|CA| \cos{\theta} \right) \\

&= |CA|^2 + |AP|\cdot \left( |AP| + |AB| \right) \\

&= |CA|^2 + |AP| \cdot|PB|
\end{align*}

etotheipi
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Wait - I didn't really read the question and at second glance I don't think that's what it's asking. And on that note, I don't actually know what it means to prove equivalence with Pythagoras...

Mentor
It means the one implies the other and vice versa.

etotheipi
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Oh, okay! In that case for the direction transversal ##\implies## Pythagoras you can just take ##P = D## (in which case ##|BP| = |AP|##). For the converse, consider first ##P \in \overline{AB}##. Then\begin{align*}
h^2 + |PD|^2 &= |CP|^2 \\
h^2 + |AD|^2 &= |CA|^2\end{align*}and so \begin{align*}

|CP|^2 &= |CA|^2 - (|AD|^2 - |PD|^2) \\

&= |CA|^2 - |AP| \cdot |BP|

\end{align*}since ##|AD| + |PD| = |BD| + |PD| = |BP|## and also ##|AD| - |PD| = |AP|##. You also take the exact same approach for when ##P \not \in \overline{AB}##.

julian
Gold Member
#10

The minimal polynomial of ##AB## is the unique monic polynomial ##p## of smallest degree such that ##p (AB) = 0##. Explicitly ##p (AB) = 0## reads:

##
a_0 \mathbb{1} + a_1 AB + a_2 (AB)^2 + \cdots + (AB)^m = 0 \quad (*)
##

Say ##A## is regular, multiplying from the left by ##A^{-1}## and from the right by ##A## gives

##
a_0 \mathbb{1} + a_1 BA + a_2 (BA)^2 + \cdots + (BA)^m = 0 ,
##

This is the smallest monic polynomial in ##BA## such that it is equal to zero. To prove this, suppose otherwise and assume that

##
q (BA) = b_0 \mathbb{1} + b_1 BA + b_2 (BA)^2 + \cdots + (BA)^l = 0 \quad (**)
##

where ##l < m##. Multiplying ##(**)## from the left by ##A## and from the right with ##A^{-1}## would result in ##q (AB) = 0##. A contradiction seeing as ##(*)## is the smallest monic polynomial in ##AB## that is equal to zero.

Is it not true as well, if ##A## is singular. Consider these two ##n \times n## matrices:

##
A =
\begin{pmatrix}
0 & \cdots & 1 \\
\vdots & & \vdots \\
0 & \cdots & 0
\end{pmatrix}
B =
\begin{pmatrix}
0 & \cdots & 0 \\
\vdots & & \vdots \\
0 & \cdots & 1
\end{pmatrix}
##

Then

##
AB =
\begin{pmatrix}
0 & \cdots & 1 \\
\vdots & & \vdots \\
0 & \cdots & 0
\end{pmatrix}
##

whereas

##
BA =
\begin{pmatrix}
0 & \cdots & 0 \\
\vdots & & \vdots \\
0 & \cdots & 0
\end{pmatrix}
##

So ##BA = 0##. We have that ##(AB)^2 = 0##. So ##BA## has the minimal polynomial is ##p (z) = z##. And ##AB## has the minimal polynomial is ##p (z) = z^2##.

Last edited:
etotheipi
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8. Let ##a_{n+1}=2+\sqrt{4+a_n}\, , \,a_0\geq -4\,,## be a sequence of real numbers. Determine - if existent - its limit in dependence of the initial value ##a_0,## and show that ##a_n\in [2,5]## in cases where ##a_0\in[-4,5],## and ##a_n\geq 5## in cases where ##a_0\geq 5## ##(n\in \mathbb{N}).##
I'll try, I'm not so confident with these so some pointers would be appreciated. It is first worth to note that if ##a_0 \geq -4## then ##a_1 \geq 2##. And if ##a_k > 0## then ##a_{k+1} = 2 + \sqrt{4+ a_k} > 0## so all the ##a_k## for ##k > 0## are positive. Now consider\begin{align*}

|a_{n} - 5| = |\sqrt{4+a_{n-1}} - 3| &= \left| \frac{(\sqrt{4+a_{n-1}} + 3)(\sqrt{4+a_{n-1}} - 3)}{(\sqrt{4+a_{n-1}} + 3)} \right| \\ \\

&= \left| \frac{a_{n-1} - 5}{\sqrt{4+a_{n-1}} + 3} \right| \\

&\leq \frac{1}{5} \left| a_{n-1} - 5 \right| \\

&\leq \dots \leq \frac{1}{5^{n-1}} \left| a_1 - 5 \right|

\end{align*}So given any ##\epsilon > 0## we can choose ##N## as any integer greater than ##1 + \mathrm{log}_5 \frac{|a_1 - 5|}{\epsilon}## in order to have ##|a_n - 5| < \epsilon## for all ##n > N##, and the limit is hence ##L = 5## for ##a_0 \in [-4, \infty)##.

For the last bit I suppose you can just say that if ##a_0 \geq -4## then ##a_1 \geq 2## and if ##a_k \geq 2## then ##a_{k+1} = 2 + \sqrt{4+a_k} \geq 2##. Similarly, if ##a_k < 5## then ##a_{k+1} = 2 + \sqrt{4 + a_k} < 2 + \sqrt{9} = 5##. So all the terms with ##k > 0## will satisfy ##a_k \in [2,5)##.

And if ##a_k \geq 5## then ##a_{k+1} = 2 + \sqrt{4 + a_k} \geq 2 + \sqrt{9} = 5##, so for ##a_0 \geq 5## you have ##a_k \geq 5## for all ##k##.

Is it okay, or are there problems?

Last edited:
Mentor
I'll try, I'm not so confident with these so some pointers would be appreciated. It is first worth to note that if ##a_0 \geq -4## then ##a_1 \geq 2##. And if ##a_k > 0## then ##a_{k+1} = 2 + \sqrt{4+ a_k} > 0## so all the ##a_k## for ##k > 0## are positive. Now consider\begin{align*}

|a_{n} - 5| = |\sqrt{4+a_{n-1}} - 3| &= \left| \frac{(\sqrt{4+a_{n-1}} + 3)(\sqrt{4+a_{n-1}} - 3)}{(\sqrt{4+a_{n-1}} + 3)} \right| \\ \\

&= \left| \frac{a_{n-1} - 5}{\sqrt{4+a_{n-1}} + 3} \right| \\

&\leq \frac{1}{5} \left| a_{n-1} - 5 \right| \\

&\leq \dots \leq \frac{1}{5^{n-1}} \left| a_1 - 5 \right|

\end{align*}So given any ##\epsilon > 0## we can choose ##N## as any integer greater than ##1 + \mathrm{log}_5 \frac{|a_1 - 5|}{\epsilon}## in order to have ##|a_n - 5| < \epsilon## for all ##n > N##, and the limit is hence ##L = 5## for ##a_0 \in [-4, \infty)##.

For the last bit I suppose you can just say that if ##a_0 \geq -4## then ##a_1 \geq 2## and if ##a_k \geq 2## then ##a_{k+1} = 2 + \sqrt{4+a_k} \geq 2##. Similarly, if ##a_k < 5## then ##a_{k+1} = 2 + \sqrt{4 + a_k} < 2 + \sqrt{9} = 5##. So all the terms with ##k > 0## will satisfy ##a_k \in [2,5)##.

And if ##a_k \geq 5## then ##a_{k+1} = 2 + \sqrt{4 + a_k} \geq 2 + \sqrt{9} = 5##, so for ##a_0 \geq 5## you have ##a_k \geq 5## for all ##k##.

Is it okay, or are there problems?
Well, a bit sloppy a.u. The proof about the ranges would have required a monotony argument and/or an induction, and for the question about the fixed point, a consideration about uniqueness would have been nice.

etotheipi
Some idea on #8...

##\int_{0}^{\pi} \frac{1}{1-\cos x} \rightarrow \int_{0}^{\pi}\frac{1+\cos x}{1-\cos^2 x} \rightarrow \int_{0}^{\pi} \csc^2 x + \frac{\cos x}{\sin^2 x}## Which looks like it diverges, so if it's not suppose to diverge, this would be the wrong way to go about it

Office_Shredder
Staff Emeritus
Gold Member
For #10, I'm confused why you specified that A is regular, it seems like the solution just uses that it's invertible. Are those the same thing?

Mentor
For #10, I'm confused why you specified that A is regular, it seems like the solution just uses that it's invertible. Are those the same thing?
It's the same in the finite-dimensional case. I have not really an idea about minimal polynomials and other properties in the infinite-dimensional case. Here it is bijective = isomorphism = invertible = non-singular = regular. At least I have learnt it as such.

etotheipi
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I was just playing around with number 9, I think if you subject the points to the following affine transformations:

- translate by ##(-1,-1)##
- then rotate by ##\frac{1}{\sqrt{17}}\begin{pmatrix} 4 & 1 \\ -1 & 4 \end{pmatrix}##
- then shear by ##\begin{pmatrix} 1 & \frac{-1}{4} \\ 0 & 1 \end{pmatrix}##
- then stretch by ##\begin{pmatrix} 1 & 0 \\ 0 & \frac{17\sqrt{3}}{36} \end{pmatrix}##

then you end up with an equilateral triangle with vertices ##\mathcal{V} \in \{ (0,0), (\sqrt{17},0), (\sqrt{17}/2, \sqrt{51}/2) \}##. The overall affine transformation is$$\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{17}}{4} & 0 \\ \frac{-\sqrt{17}}{12\sqrt{3}} & \frac{\sqrt{17}}{3\sqrt{3}} \end{pmatrix} \begin{pmatrix} x-1 \\ y-1 \end{pmatrix}$$In the resulting equilateral triangle the maximal inscribed ellipse is just a circle of centre ##\mathbf{R}_C = (\sqrt{17}/2, \sqrt{51}/ 6)## and radius ##\mathscr{R} = \sqrt{51}/6##. The area of the original ellipse should simply scale as ##A_{\mathrm{ellipse}} = (\mathrm{det} \mathcal{M}^{-1})^2 A_{\mathrm{circle}} = \frac{1}{(\mathrm{det} \mathcal{M})^2} A_{\mathrm{circle}}##.

Also, the circle has equation ##\left(X - \frac{\sqrt{17}}{2} \right)^2 + \left(Y - \frac{\sqrt{51}}{6} \right)^2 = \frac{51}{36}##, so we can just substitute in for ##X(x,y)## and ##Y(x,y)## to find the equation of the ellipse.

I'm a bit tired to do remainder of the algebra now, though!

Last edited:
hutchphd and Infrared
Mentor
I want to see the center coordinates (up to 2 digits), foci coordinates (up to 2 digits), semi-axis (exact), area (exact). I need them to check whether it makes sense to read anything. No translated, rotated, or otherwise manipulated orbit equations. I am too lazy to decode the information from implicitly given equations.

Infrared
Gold Member
##\int_{0}^{\pi} \frac{1}{1-\cos x} \rightarrow \int_{0}^{\pi}\frac{1+\cos x}{1-\cos^2 x} \rightarrow \int_{0}^{\pi} \csc^2 x + \frac{\cos x}{\sin^2 x}## Which looks like it diverges, so if it's not suppose to diverge, this would be the wrong way to go about it

Your integral diverges but the integral in the problem converges. A similar example is that ##\int_0^1\frac{1}{x^2}dx## diverges but ##\iiint_{x^2+y^2+z^2\leq 1}\frac{1}{x^2+y^2+z^2}dx dy dz## converges (use spherical coordinates).

romsofia
etotheipi
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I want to see the center coordinates (up to 2 digits), foci coordinates (up to 2 digits), semi-axis (exact), area (exact). I need them to check whether it makes sense to read anything. No translated, rotated, or otherwise manipulated orbit equations. I am too lazy to decode the information from implicitly given equations.
That said... my ellipse is:$$F(x,y) := \frac{119}{108}x^2 -\frac{17}{54} xy + \frac{17}{27}y^2 -\frac{17}{3}x - \frac{17}{6}y + \frac{34}{3} = 0$$the centre can be found either by the inverse affine map from in #12 or simply by setting ##\nabla F(x_c, y_c) = 0## which gives ##(x_c, y_c) = (3,3)##. Is that what you got, as a start?

Mentor
I haven't calculated the ellipse. The center is ##(3,3)##, yes.

etotheipi
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I have for the semi-major, semi-minor and area:\begin{align*}

a &= \frac{\sqrt{2}(11+ \sqrt{13})\sqrt{11-\sqrt{13}}}{36} \\

b &= \frac{\sqrt{2}(11- \sqrt{13})\sqrt{11+\sqrt{13}}}{36} \\

A_{\mathrm{ellipse}} &= \pi ab\end{align*}the centre-to-focus length is ##c = \sqrt{a^2 - b^2}##. The focal points are then the ##(x_{f\pm}, y_{f\pm}) = (3 \pm c\cos{\theta}, 3 \pm c\sin{\theta})## where ##\cos{\theta}## and ##\sin{\theta}## satisfy\begin{align*}

\frac{119}{108} &= a^2 \sin^2{\theta} + b^2\cos^2{\theta} = (a^2 - b^2)\sin^2{\theta} + b^2

\end{align*}which could be found with a calculator, I guess

Last edited:
hutchphd
etotheipi
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Do you have any hints for the integral in problem 3? I could only tell so far that with ##x \mapsto u(x) = \pi - x## you see ##\mathcal{I} = \int_{\mathcal{D}} \frac{\mathrm{d}^3 x}{1-c(x)c(y)c(z) } = \int_{\mathcal{D}} \frac{\mathrm{d}^3 x}{1+c(x)c(y)c(z) } = \int_{\mathcal{D}} \frac{\mathrm{d}^3 x}{1-c(x)^2c(y)^2c(z)^2 }##...

Mentor
I have for the semi-major, semi-minor and area:\begin{align*}

a &= \frac{\sqrt{2}(11+ \sqrt{13})\sqrt{11-\sqrt{13}}}{36} \\

b &= \frac{\sqrt{2}(11- \sqrt{13})\sqrt{11+\sqrt{13}}}{36} \\

A_{\mathrm{ellipse}} &= \pi ab\end{align*}the centre-to-focus length is ##c = \sqrt{a^2 - b^2}##. The focal points are then the ##(x_{f\pm}, y_{f\pm}) = (3 \pm c\cos{\theta}, 3 \pm c\sin{\theta})## where ##\cos{\theta}## and ##\sin{\theta}## satisfy\begin{align*}

\frac{119}{108} &= a^2 \sin^2{\theta} + b^2\cos^2{\theta} = (a^2 - b^2)\sin^2{\theta} + b^2

\end{align*}which could be found with a calculator, I guess
I have different values, e.g. no ##\sqrt{13}.## And the area is a nice number, not a monster. You still deliver implicit solutions, instead of good old digits. However, I used a different method for calculation, so it's hard to compare them, especially as you didn't post what you had done exactly. E.g. the optimization process remains in the dark.

etotheipi
Mentor
Do you have any hints for the integral in problem 3? I could only tell so far that with ##x \mapsto u(x) = \pi - x## you see ##\mathcal{I} = \int_{\mathcal{D}} \frac{\mathrm{d}^3 x}{1-c(x)c(y)c(z) } = \int_{\mathcal{D}} \frac{\mathrm{d}^3 x}{1+c(x)c(y)c(z) } = \int_{\mathcal{D}} \frac{\mathrm{d}^3 x}{1-c(x)^2c(y)^2c(z)^2 }##...
@Infrared already gave a hint. Weierstraß is another.

etotheipi
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And the area is a nice number, not a monster.
When I simplified my expression for ##A_{\mathrm{ellipse}}## using the exact expressions for ##a \sim 1.56## and ##b \sim 1.11## in post #19 I get ##A_{\mathrm{ellipse}} = \pi\sqrt{3}##. I didn't calculate the focal coordinates exactly but I have ##\boldsymbol{f}_{+} \sim (3.32, 4.05)## and ##\boldsymbol{f}_{-} \sim (2.68, 1.95)##

Last edited:
etotheipi
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@Infrared already gave a hint. Weierstraß is another.
Hmm, okay I still don't know so I'll wait and see how someone else solves it! Weierstrass takes it to ##\mathcal{I} = 4\int_{\mathcal{D}'} (t^2u^2v^2 + t^2 + u^2 + v^2)^{-1} \, \mathrm{d} t \, \mathrm{d}u \, \mathrm{d}v## with ##\mathcal{D}' = (0, \infty)^{3}## but it also looks difficult to solve.

Mentor
When I simplified my expression for ##A_{\mathrm{ellipse}}## using the exact expressions for ##a \sim 1.56## and ##b \sim 1.11## in post #19 I get ##A_{\mathrm{ellipse}} = \sqrt{3}##. I didn't calculate the focal coordinates exactly but I have ##\boldsymbol{f}_{+} \sim (3.32, 4.05)## and ##\boldsymbol{f}_{-} \sim (2.68, 1.95)##
You drive me crazy: sloppy a.u.! You dropped the ##\pi## in the area. The numbers are correct, except that I haven't checked ##a,b##. Anyway, since we now have a collection of posts with partial solutions, I will post mine for the sake of completeness, and in case someone wants to learn about the Steiner ellipse:

Set
\begin{align*}
\mathbb{C}[x]\ni p(z)&=(z-1-i)(z-5-2i)(z-3-6i)\\&=z^3 - (9 + 9 i) z^2 + (3 + 52 i) z + (33 - 39 i)
\end{align*}
The maximal inscribed ellipse of the triangle of zeros of ##p(x)## is thus the Steiner inellipse, where the sides of the triangle are tangents at their midpoints. The foci are the zeros of ##p'(z)## and the center the zero of ##p''(x)## by Marden's theorem.
\begin{align*}
0&=p'(z)=3 z^2 - (18 + 18 i) z + (3 + 52 i)\\
&=3\left(z- (3 + 3 i) - \sqrt{-1+\dfrac{2i}{3}}\right)\left(z- (3 + 3 i)+\sqrt{-1+\dfrac{2i}{3}}\right)\\
&\approx 3\left(z-3.32-4.05 i \right)\left(z-2.68 - 1.95 i \right)\\
0&=p''(x)=6\left(z-(3+3i)\right)
\end{align*}
Hence the center of the ellipse is ##(3,3)## and the foci ##(2.68,1.95),(3.32,4.05).## The area of an ellipse is given as ##A=\pi ab ,## so we have to compute the semi-axis of a Steiner inellipse within ##\triangle ABC## with center ##S.##
\begin{align*}
A=(1,1)\, , \,B=(5,2)\, &, \,C=(3,6)\, , \,S=(3,3)\, , \,\\
M:=\dfrac{1}{4}\left(\overline{SC}^2+\dfrac{1}{3}\overline{AB}^2\right)\, &, \,N:=\dfrac{1}{4\sqrt{3}}\cdot \left|\det\left(\vec{SC},\vec{AB}\right)\right|\\
a=\dfrac{1}{2}\left(\sqrt{M+2N}+\sqrt{M-2N}\right)\, &, \,b=\dfrac{1}{2}\left(\sqrt{M+2N}-\sqrt{M-2N}\right)
\end{align*}
\begin{align*}
M&=\dfrac{1}{4}\left(9+\dfrac{1}{3}\cdot 17\right)=\dfrac{11}{3}\\
N&=\dfrac{1}{4\sqrt{3}}\left|\det\left(\begin{bmatrix}
0&4\\3&1\end{bmatrix}\right)\right|=\sqrt{3}\\
a&=\dfrac{1}{2\sqrt{3}}\left(\sqrt{11+6\sqrt{3}}+\sqrt{11-6\sqrt{3}}\right)\\
b&=\dfrac{1}{2\sqrt{3}}\left(\sqrt{11+6\sqrt{3}}-\sqrt{11-6\sqrt{3}}\right)\\
ab&=\dfrac{1}{12}\left(\left(11+6\sqrt{3}\right)-\left(11-6\sqrt{3}\right)\right)=\sqrt{3}\\
\end{align*}

etotheipi
benorin
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@fresh_42 does the answer to #1) involves ##\Gamma \left( \tfrac{7}{3}\right)##? I got that times a divergent integral but the calculus I did was slapdash, just want to know if I’m out on the correct limb here.

Mentor
@fresh_42 does the answer to #1) involves ##\Gamma \left( \tfrac{7}{3}\right)##? I got that times a divergent integral but the calculus I did was slapdash, just want to know if I’m out on the correct limb here.
I have ##1## and ##2## as numerator, not ##7##, hence the answer is maybe.