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Acute triangle $ABC$ has side length $a \ge b \ge c$. We drop the altitude from point $C$ to side $c$, and let its length be h. We construct a square with side length $h$ so that side $c$, points $A$ and $C$ lie on the square's sides and point $B$ lies on its vertex. Then, we construct another square with side $s$ so that $C$ lies on its vertex and points $A$ and $B$ lie on its sides. Which square is bigger?

I found the equations $c^2=a^2+b^2-2s\sqrt{a^2-s^2}-2s\sqrt{b^2-s^2}$ and $4c^2h^2=a^4-2a^2b^2-2a^2c^2+b^4+6b^2c^2+c^4$

TNT1288
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2 Answers2

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This problem is underconstrained. In the square with side $s$ let any qualifying triangle be constructed. Then $h=s$ iff the foot of the altitude to $C$ is on the quarter circle with radius $s$ centered on $C$. And we can clearly construct triangles with feet inside, on, or outside that arc. Your example is outside.

Edit: added picture illustrating 3 possibilities, right panel is equilateral.

enter image description here

RobinSparrow
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  • I don't understand your reasoning. $s$ is unique, are you trying to claim otherwise? – D S Mar 15 '24 at 09:16
  • The problem is simpler in reverse. Fix the second square first, in it construct this or any other acute triangle with vertex $C$ and the other vertices on top and left side. Then $h$ is in the square yet it may be greater than, equal to, or less than the side $s$. Consider the isosceles constructs with a near $90$ apex angle $C$ or near $0$ angle for $h$ less than and greater than $s$, respectively. – RobinSparrow Mar 15 '24 at 12:28
  • Your construction is vague, but from what I understand, you haven't used $a \ge b \ge c$. – D S Mar 15 '24 at 12:50
  • I’m not seeing the problem or the vagueness, I said let it be any qualifying triangle per the initial description. And the ones I just described qualify. – RobinSparrow Mar 15 '24 at 12:55
  • Can you add a diagram with $h<s$? – D S Mar 15 '24 at 13:13
  • @DS picture added. In my pre-coffee fuddle this am I mentioned a near-right isosceles in my comment response, that's not qualifying, I forgot the constraints overnight, the equilateral above is the case I should have said there. – RobinSparrow Mar 15 '24 at 13:58
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    Thanks, it is clear now – D S Mar 15 '24 at 14:04
  • would there be any restrictions on $a$, $b$ and $c$ that would classify any triangle in each of the three categories? – TNT1288 Mar 15 '24 at 21:22
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    You could definitely generate the function but you’d need to parameterize the situation. I would slide A along the top from left to right, which restrains B to values yielding longer sides than A yet not more than 90 degrees, and calculate where on that interval you have tangency. The solution is a surface over the 2 parameters. Left as an exercise for the reader – RobinSparrow Mar 15 '24 at 22:12
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In the attached figure, your second triangle is red color and your first one is the same black color. You have $$\begin{cases}a\cos(t)=s\\a\cos(u)=h\end{cases}\Rightarrow \frac{s}{h}=\frac{\cos(t)}{\cos(u)}$$ Since $\triangle{ABC}$ is acute we have three possibilities $$ s\lt h\iff \cos(t)\lt\cos(u)\iff t\gt u\\s\gt h\iff \cos(t)\gt\cos(u)\iff t\lt u\\s=h\iff t=u$$ Obviously these possibilities mean respectively black is bigger,red is bigger and both are equal.

In the figure there are particular values to verify the size of concerned squares with $(a,b,c)=(10,8,7)$ and the literal calculation is similar but more difficult to manipulate.

We have $h\approx7.946$; the value of $s$ is not hard to determine; we have the system of three unknowns in which we eliminate $y$ and $z$ $$ y^2+z^2=7^2\\s^2+(s-z)^2=8^2\\s^2+(s-y)^2=10^2$$ We get $s\approx7.91643$ and $s\approx 3.4658$ so the black square is bigger.

Note the values of $h$ and $s$ are near. There are positions for $s\lt h$ and $s\gt h$ for which it is impossible the construction of squares such as asked but I not explain about it.

enter image description here

Piquito
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