IMO selected problem

Geometry Level 5

F ( x , y ) = ( x + 1 ) 2 + ( y − 1 ) 2 + ( x − 1 ) 2 + ( y + 1 ) 2 + ( x + 2 ) 2 + ( y + 2 ) 2 F(x,y)=\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x-1)^2+(y+1)^2}+\sqrt{(x+2)^2+(y+2)^2} Let the minimum of the expression above is S S . If S 2 = k + m n S^{2}=k+m\sqrt{n} , where n n is square free, find m + k + n m+k+n .


The answer is 25.

Son Nguyen by Son Nguyen , 20, Vietnam

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2 solutions

Fletcher Mattox
Aug 22, 2020

since we haven't a solution among 14 solvers, i'll offer this hack: 

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from util import realrange
from math import sqrt

def f(x, y):
    return sqrt((x+1)**2 + (y-1)**2) + sqrt((x-1)**2 + (y+1)**2) + sqrt((x+2)**2 + (y+2)**2)

def g(m, k, n):
    return k + m * sqrt(n)

fmin = 10**6
for a in realrange(-.58, -.57, 1000):
    for b in realrange(-.58, -.57, 1000):
        t = f(a, b)
        if t < fmin:
            fmin = t 

target = fmin**2
for m in range(1, 20):
    for k in range(1, 20):
        for n in range(1, 10):
            if abs(g(m, k, n) - target) < 10**-10:
                print(m + k + n)


1
 
25

1 Helpful 1 Interesting 1 Brilliant 1 Confused
Ajit Athle
Dec 20, 2020

The given expression is the sum of distances from triangle vertices (-1,1), (-2,-2) & (1,-1) from the Fermat Point of the triangle. If x, y & z be these distances then, x²+y²+xy=10, z²+y²+zy=10, x²+z²+xz=8 and s²=(x+y+z)^2. We obtain, s² = 14+8√3

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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