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Algebra Level 4

Three positive real numbers satisfy the following: $\begin{aligned} x^2+xy+y^2 & = 9 \\ y^2+yz+z^2 & = 16 \\ z^2+xz+x^2 & = 25 \end{aligned}$ If $xy+yz+xz=a\sqrt{b}$ , where $a$ and $b$ are integers and $b$ is square-free, then find the value of $a+b$ .

The answer is 11.

2 solutions

Ajit Athle
Dec 10, 2020

Consider right triangle ABC with AB=3, BC=4 & CA=5. Let F be the Femat point of the triangle and let AF=x, BF=y and CF=z. Then, <AFB, <BFC & <CFA are all 120°. And thus by the co-sine law in triangles, AFB, BFC & CFA, we claim x²+y²+xy=AB²=9. Likewise, y²+z²+yz=BC²=16 and z²+x²+zx=CA²=25. Now A(Tr ABC) = A(Tr.AFB) + A(Tr.BFC) + A(CFA) or xy √3/4 + yz √3/4 + zx √3/4 = 3 4/2 = 6 or xy + yz + zx = 6*4/√3 = 8√3 or a+b = 11

Fletcher Mattox
Nov 23, 2020

I believe you should require positive integers, not integers, as there seems to be a negative solution.

from sympy import *
from sympy.abc import x, y, z

eq1 = x**2 + x*y + y**2 - 9
eq2 = y**2 + y*z + z**2 - 16
eq3 = z**2 + x*z + x**2 - 25

sol = solve([eq1, eq2, eq3], [x, y, z])
for x, y, z in sol:
    t = x*y + y*z + x*z
    print(simplify(t))

8*sqrt(3)
8*sqrt(3)
-8*sqrt(3)
-8*sqrt(3)

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The answer is 11.

2 solutions

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