MHB Triangle Lengths: Can $a,b,c$ Form a Triangle?

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The equation given, $a^2 + 5b^2 + 4c^2 - 4ab - 4bc = 0$, can be analyzed to determine if the values of $a$, $b$, and $c$ can represent the sides of a triangle. By applying the triangle inequality, the conditions $a + b > c$, $a + c > b$, and $b + c > a$ must be satisfied. The equation can be rearranged and factored to explore the relationships between the variables. Ultimately, a solution must demonstrate that the values derived from the equation fulfill the triangle inequality criteria. Therefore, the conclusion hinges on whether valid positive values for $a$, $b$, and $c$ exist that satisfy both the equation and the triangle inequalities.
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Three positive real numbers $a,\,b$ and $c$ are such that $a^2+5b^2+4c^2-4ab-4bc=0$. Can $a,\,b$ and $c$ be the lengths of the sides of a triangle? Justify your answer.
 
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We have

$a^2+5b^2+4c^2-4ab-4bc=0$

Or $a^2-4ab + 4b^2 + b^2 - 4bc + 4c^2 = 0$

Or $(a-2b)^2 + (b- 2c)^2= 0$

Because a,b,c are real so each part is zero

This gives the solution a = 2b and b = 4c

Or a = 4c, b =2c and so $a > b+ c$

so the answer is no
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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