Three squares are shown in the diagram. The largest has side $AB$ of length $1$. The others have side $AC$ of length $x$, and side $DE$ of length $y$. As $D$ moves along $AB$, the values of $x$ and $y$ change. Determine the values of $x$ and $y$ for which $x^2 + y^2$ is a minimum. What is this minimum?
On
Hint: Let $a:=AD-\dfrac{AB}{2}$, so $a\in\left[-\dfrac12,+\dfrac12\right]$. Prove that $x=\left(\dfrac12+a\right)\left(\dfrac12-a\right)=\dfrac14-a^2$ and $y=\sqrt{\left(\dfrac12+a\right)^2+\left(\dfrac12-a\right)^2}=\sqrt{\dfrac12+2a^2}$. Then, $$x^2+y^2=\left(\frac14-a^2\right)^2+\left(\dfrac12+2a^2\right)=\frac{9}{16}+\frac{3}{2}\,a^2+a^4\,.$$
You should immediately get that $$\dfrac9{16}\leq x^2+y^2\leq 1\,.$$
On
Let $ACFH$ be a square inside the triangle $GAD$ (See your picture).
Thus, $AH=HF=FC=AC=x$ and $GD=y$.
Also, let $AG=a$ and $AD=b$.
Hence, $a+b=1$, $a^2+b^2=y^2$ and $$\frac{HF}{AD}=\frac{GH}{AG}$$ or $$\frac{x}{b}=\frac{a-x}{a}$$ or $$x\left(\frac{1}{a}+\frac{1}{b}\right)=1$$ or $$x=ab.$$ Now, by AM-GM $$ab\leq\left(\frac{a+b}{2}\right)^2=\frac{1}{4}$$ and $$x^2+y^2=a^2b^2+a^2+b^2=a^2b^2-2ab+(a+b)^2=ab(ab-2)+1\geq\frac{1}{4}\left(\frac{1}{4}-2\right)+1=\frac{9}{16}.$$ The equality occurs for $a=b=\frac{1}{2},$ which says that we got a minimal value.
Id est, $$x=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$$ and $$y=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\frac{1}{\sqrt2}.$$
On
Let $z = AD$. Let $G$ be the vertex of the inner square opposite of $E$.
We have $AG = 1-z$. The triangles $ADG$ and $CDF$ are similar and $CF = x$ so $CD = \frac{z}{1-z}x$.
From Pythagoras we get $y^2 = AG^2 + AD^2 = z^2 + (1-z)^2$, and also $$z = AD = AC + CD = x + \frac{z}{1-z}x$$
which gives $x = z(1-z)$.
Therefore $$x^2 + y^2 = z^2 + (1-z)^2 + z^2(1-z)^2 = z^2(1-z)^2 + 1 - 2z(1-z) = z(1-z)[z(1-z) - 2] + 1$$
Since $z \in [0,1]$ we have $z(1-z) \in \left[0,\frac14\right]$ and $z \mapsto z(z-2)$ is decreasing on $\left[0,\frac14\right]$ so
$$x^2 + y^2 = z(1-z)[z(1-z) - 2] + 1 \ge \frac14\left[\frac14 - 2\right] + 1 = \frac{9}{16}$$
the minimum value being attained for $z = \frac12$, meaning $x = \frac14$ and $y = \frac1{\sqrt{2}}$.
On
The obvious bounds are: $$0\le x^2\le \frac1{16}; \ \ \ \frac12\le y^2\le 1.$$ However, the extremes can not be added, because $x$ and $y$ are oppositely related (when $x$ is min, $y$ is max and vice versa). In fact: $$\frac1{16}+\frac12\le x^2+y^2\le 0+1$$ However, the extremes may happen in between. So, let's find the relationship between $x$ and $y$.
Label $CD=z$.
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From similarity of $\Delta CDF$ and $\Delta BDE$: $$\frac{CF}{BD}=\frac{CD}{BE} \Rightarrow \frac{x}{1-x-z}=\frac{z}{x+z} \Rightarrow z=(x+z)^2 \tag{1}$$ From the Pythagorean Theorem: $$\begin{align}DE^2&=BD^2+BE^2 \Rightarrow \\ y^2&=(1-x-z)^2+(x+z)^2=1-2(x+z)+2(x+z)^2=\\ &=1-2(x+z)+2z=\\ &=1-2x \end{align}$$ Noting the true bound $0\le x\le \frac14$, we get: $$x^2+y^2=x^2+1-2x=(x-1)^2\ge \frac9{16},$$ and the equality occurs for the maximum $x=\frac14$.
Assume $\angle DFC=\theta, \theta \in \left(0,\dfrac{\pi}{2}\right)$. Then $$1=AB=AD+DB=y\sin\theta+y\cos \theta.$$ Thus $$y=\frac{1}{\sin \theta+\cos \theta}.$$ Moreover, by the equality on the areas of triangles, we have $$\frac{1}{2}\cdot y\sin \theta \cdot x+\frac{1}{2}\cdot y\cos \theta \cdot x=\frac{1}{2}\cdot y\sin \theta \cdot y\cos \theta,$$ namely $$x=\frac{y\sin \theta \cos \theta}{\sin \theta+\cos \theta}=\frac{\sin\theta\cos\theta}{(\sin \theta+\cos \theta)^2}.$$Therefore, $$x^2+y^2=\frac{\sin^2\theta\cos^2\theta}{(\sin \theta+\cos \theta)^4}+\frac{1}{(\sin \theta+\cos \theta)^2}.$$ Dnote $\sin \theta+\cos \theta=z, z \in (1,\sqrt{2}] .$ Then $\sin \theta \cos \theta=\dfrac{z^2-1}{2}.$Thus $$x^2+y^2=\frac{1}{4}\left(1+\frac{1}{z^2}\right)^2\geq \frac{1}{4}\left(1+\frac{1}{(\sqrt{2})^2}\right)^2=\frac{9}{16},$$with the equality holding if and only if $z=\sqrt{2}$, namely, $\theta=\dfrac{\pi}{4}$, which implies $x=\dfrac{1}{4}$ and $y=\dfrac{\sqrt{2}}{2}.$