Triangle area maximized

Geometry Level 3

Let a a , b b , and c c be the side lengths of A B C \triangle ABC and 2 a + 7 b + 11 c = 120 2a+7b+11c=120 . If the maximum area of A B C \triangle ABC is equal to m n m\sqrt{n} , where m m and n n are integers with n n square-free, find the value of m + n m+n .


The answer is 13.

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2 solutions

Chris Lewis
Nov 26, 2020

For neatness, we'll maximise 16 16 times the square of the area first. Using Heron's formula, this is ( a + b + c ) ( a + b c ) ( a b + c ) ( a + b + c ) (a+b+c)(a+b-c)(a-b+c)(-a+b+c)

We can now include the constraint using a Lagrangian multiplier; so define

F ( a , b , c , L ) = ( a + b + c ) ( a + b c ) ( a b + c ) ( a + b + c ) + L ( 2 a + 7 b + 11 c 120 ) F(a,b,c,L)=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)+L(2a+7b+11c-120)

Setting each of the partial derivatives to zero, 4 a ( a 2 + b 2 + c 2 ) + 2 L = 0 4 b ( a 2 b 2 + c 2 ) + 7 L = 0 4 c ( a 2 + b 2 c 2 ) + 11 L = 0 2 a + 7 b + 11 c 120 = 0 \begin{aligned} 4a \left(-a^2+b^2+c^2 \right)+2L &= 0 \\ 4b \left(a^2-b^2+c^2 \right)+7L &= 0 \\ 4c \left(a^2+b^2-c^2 \right)+11L &= 0 \\ 2a+7b+11c-120 &=0 \end{aligned}

Solving this (ie, getting Wolfram|Alpha to solve it) we find just one solution with positive lengths, a = 5 a=5 , b = 7 b=7 and c = 8 c=8 giving an area of 10 3 10\sqrt3 and an answer of 13 \boxed{13} .

Hi @Chris Lewis , we have posted your report as a solution.

Brilliant Mathematics Staff - 6 months, 2 weeks ago
Fletcher Mattox
Nov 24, 2020
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from scipy.optimize import minimize
from sympy import *

def area(x):
    a, b, c = x
    s = (a + b + c)/2
    return -sqrt(s*(s - a)*(s - b)*(s - c))

def constraint(x):
    a, b, c = x
    return 2*a + 7*b + 11*c - 120

cons=({'type':'eq','fun':constraint})
x0 = [10, 10, 10]
res= minimize(area,x0,method='SLSQP',constraints=cons)
a, b, c = list(map(Integer, list(map(round, res.x))))
print(-area([a, b, c]))

1
10*sqrt(3)

How do you know a , b , c a,b,c are integers? How do you know the triangle form must be a non-degenerate triangle?

If you change the number 120 to 60, your code shows an error message, even though the maximum area of the triangle occurs when ( a , b , c ) = ( 4 , 7 2 , 5 2 ) . (a,b,c) = (4, \tfrac72, \tfrac52) . So your code is incomplete.

Pi Han Goh - 6 months, 2 weeks ago

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Yes, the code is a hack. I'm learning optimisation and thought it was a cute hack. As I typed it, I actually anticipated you would call me out for it. I deserve it!

Fletcher Mattox - 6 months, 2 weeks ago

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