Yet another 3-4-5 triangle

Geometry Level 5

Inspiration , along with a few other questions from and inspired by @Fletcher Mattox

Given a point $Q$ , a triangle $\Delta ABC$ is constructed so that $AQ=3$ , $BQ=4$ , $CQ=5$ , as shown in the diagram above. Among all such triangles, what is the largest possible perimeter?

If this perimeter is $p$ , enter $\left\lfloor 1000 p \right\rfloor$ as your answer.

Bonus: Can you find a closed form for $p$ ?

The answer is 20932.

1 solution

Fletcher Mattox
May 13, 2021

I generally don't post code for these types of problems, but Chris egged me on. :)

here is a picture of the triangle.

from math import sin, cos, degrees, radians, pi
from sympy import *
from scipy.optimize import minimize
import numpy
numpy.set_printoptions(precision=15)

# let the arbitrary point be the origin
# parameters to doit() are the three angles the vertices make with the origin

def doit(x):
    A_,B_,C_ = x
    A = Point(a*cos(A_), a*sin(A_))
    B = Point(b*cos(B_), b*sin(B_))
    C = Point(c*cos(C_), c*sin(C_))
    t = Triangle(A, B, C)
    return -t.perimeter

a,b,c = 3,4,5
x0 = [radians(0), radians(120), radians(240)]
bounds = [(0, 2*pi)]*3
options = {'ftol':1e-16, 'maxiter':2000}
res = minimize(doit, x0, method='SLSQP', bounds=bounds, options=options)
print("Solution (angles with the x-axis)", list(map(degrees, res.x)))
print("Maximum perimeter: ", -N(res.fun))

1 2	`Maximum perimeter: 20.9323303679694 Solution (angles with the x-axis) [1.0277511544426605, 113.36387173594606, 242.66520311471177]`

Can you tell sympy to get a closed form of $p$ ?

It is the largest real root of $\num{3600} X^6 - \num{1568639} X^4 - \num{3869200} X^2 + \num{16257024 }= 0 .$

Pi Han Goh - 4 weeks, 1 day ago

Not that I am aware of, but apparently Mathematica can. However, I cannot afford that software.

Fletcher Mattox - 4 weeks, 1 day ago

Pi Han Goh - 4 weeks, 1 day ago

@Pi Han Goh – Alternatively via the free WolframAlpha

Pi Han Goh - 4 weeks ago

@Pi Han Goh – Let $X$ denote the angle (in radian) sandwiched between the given lengths 4 and 5.

And let $Y$ denote the angle (in radian) sandwiched between the given lengths 3 and 5.

Then we have $X,Y \in (0, \pi )$ and $X + Y > \pi.$

The the perimeter is maximized when $\left( \dfrac{4 \sin X}{3\sin Y} \right)^2 = \dfrac{41-40\cos X}{34 - 30\cos Y}$

or when $(X,Y) \approx (2.25673, 2.06582)$

Pi Han Goh - 4 weeks ago

@Pi Han Goh – Alternatively,

let $l$ denote the other length formed from the interior triangle with two given sides 3 and 4
let $m$ denote the other length formed from the interior triangle with two given sides 4 and 5
let $n$ denote the other length formed from the interior triangle with two given sides 3 and 5

The perimeter is maximized when $l^2 + \dfrac{49}{l^2} + 32 = m^2 + \dfrac{81}{m^2} = n^2 + \dfrac{256}{n^2} + 14$

or when $(l,m,n) \approx (5.84131, 8.14498, 6.94624)$

Pi Han Goh - 4 weeks ago

Thank you both for sharing your results! @Fletcher Mattox - that's very similar to my approach; I used two variables to represent angles at the centre (essentially your angles B and C; I fixed A=0, because rotating the whole triangle doesn't affect its perimeter), then equated the partial derivatives to zero to find the maximum.

@Pi Han Goh - I went about this numerically (I thought I got close to an analytical solution at one point, but no). That's a wonderful equation you found! I wonder how the coefficients come from $3,4,5$ ?

Chris Lewis - 4 weeks, 1 day ago

@Fletcher Mattox and @Chris Lewis

I've posted another problem that explains how I got the results to my 2 latest comments above

Pi Han Goh - 4 weeks ago

Uh-oh...ummm...have we all missed something here? (If so, I feel incredibly stupid for having posted this, because the inspiration problem is actually the solution!!)

Let $\angle BQC=\alpha$ , $\angle CQA=\beta$ , $\angle AQB=\gamma$ , so that $\alpha+\beta+\gamma=2\pi$ . By the cosine rule,

$\begin{aligned} a^2 &=BQ^2 + CQ^2 - 2BQ \cdot CQ \cos \alpha \\ b^2 &=CQ^2 + AQ^2 - 2CQ \cdot AQ \cos \beta \\ c^2 &=AQ^2 + BQ^2 - 2AQ \cdot BQ \cos \gamma \end{aligned}$

Now, define $F(\alpha,\beta,\gamma,L)=\sqrt{BQ^2 + CQ^2 - 2BQ \cdot CQ \cos \alpha}+\sqrt{CQ^2 + AQ^2 - 2CQ \cdot AQ \cos \beta}+\sqrt{AQ^2 + BQ^2 - 2AQ \cdot BQ \cos \gamma}+L(\alpha+\beta+\gamma-2\pi)$

(ie set it up as a Lagrangian multiplier problem). Then the partial derivative with respect to $\alpha$ is $F_\alpha=\frac{BQ \cdot CQ \sin \alpha}{\sqrt{BQ^2 + CQ^2 - 2BQ \cdot CQ \cos \alpha}}+L$

Equating this and the other partial derivatives to zero (to find the maximum) gives $\frac{BQ \cdot CQ \sin \alpha}{\sqrt{BQ^2 + CQ^2 - 2BQ \cdot CQ \cos \alpha}}=\frac{CQ \cdot AQ \sin \beta}{\sqrt{CQ^2 + AQ^2 - 2CQ \cdot AQ \cos \beta}}=\frac{AQ \cdot BQ \sin \gamma}{\sqrt{AQ^2 + BQ^2 - 2AQ \cdot BQ \cos \gamma}}=-L$

We can make two substitutions here; the denominators are just the sidelengths $a,b,c$ ; the numerators are twice the areas of the triangles $\Delta BQC$ , $\Delta CQA$ , $\Delta AQB$ respectively. So the equation becomes

$\frac{2[\Delta BQC]}{a}=\frac{2[\Delta CQA]}{b}=\frac{2[\Delta AQB]}{c}$

But these are just the heights of the three triangles $\Delta BQC$ , $\Delta CQA$ , $\Delta AQB$ . In other words, the perpendicular distances from $Q$ to the three sides are all the same. In other words, $Q$ is the incentre!

I think I went for numerical methods too quickly with the last one and thought the incentre didn't actually maximise the perimeter.

Have I made a mistake here? Or should I convert this to a solution and be very embarrassed?

Chris Lewis - 3 weeks, 5 days ago

Did you see this ?

Fletcher Mattox - 3 weeks, 4 days ago

@Fletcher Mattox – I hadn't seen it - but wish I had!

Chris Lewis - 3 weeks, 4 days ago

Let point C is on an ellipse with foci A and B, then AC+BC=major axis. From the drawing is clear, that for a fixed AB, a, b, and c, the maximum AC+BC is when the ellipse is tangent to the blue circle (Q, c), because the red ellipse has the biggest possible major axis. From the focus-to-focus reflection property => CQ is an angle bisector of an angle C. Analogically AB+AC is maximal when AQ is an angle bisector of an angle A and AB+BC is maximal when BQ is an angle bisector of an angle B => the largest possible perimeter is when Q is the intersection point of the angle bisectors of ABC or Q is the center of the incircle of ABC. From the Heron's formula => $A_{ABC}=\frac{pr}{2}=\sqrt{\frac{p}{2}xyz}\\ \frac{p}{2}=x+y+z\\x^2+r^2=a^2\\y^2+r^2=b^2\\ z^2+r^2=c^2$ If a=3, b=4, c=5 and solving numerically for p $\Rightarrow\;\;p=20.9323303679693379582020526\;\;\Rightarrow\;\; \boxed{20932}$

Iliya Hristov - 2 days, 23 hours ago

Wow, nice approach - you might want to post it as a solution rather than a comment, though (so more people see it)

Chris Lewis - 2 days, 12 hours ago

Yet another 3-4-5 triangle

The answer is 20932.

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