## SUBJECT 3

Consider the squares $\mathrm{ABCD}$ and $\mathrm{ABEF}$ with $\mathrm{AB}=1 \mathrm{~m}$, such that $\mathrm{EC}=\sqrt{2} \mathrm{~m}$. Let $\mathcal{C}(\mathrm{O}, r)$ be the circle circumscribed around quadrilateral CDFE, and draw the diameter HI of the circle, where H, I $\in \mathcal{C}(\mathrm{O}, r)$, $\mathrm{HI} \perp \mathrm{CD}$, and $\mathrm{OI} \cap \mathrm{CD}=\{\mathrm{M}\}$. If $\mathrm{HG} \perp(\mathrm{ABC})$ and $\mathrm{IQ} \perp(\mathrm{ABC})$, where $\mathrm{G}, \mathrm{Q} \in(\mathrm{ABC})$, determine the length of the radius $r$ and the area of the polygon AGBCQD.

Prof. Mihai Pîrvu
SUBIECTUL 3

If $\mathrm{ABCD}$ and $\mathrm{ABEF}$ are squares with $\mathrm{AB}=1 \mathrm{~m} \Rightarrow \mathrm{AB}=\mathrm{EF}=\mathrm{CD}=1 \mathrm{~m}, \mathrm{AB}|| \mathrm{EF}|| \mathrm{CD}$ and $\mathrm{AB} \perp(\mathrm{BCE})$. (1) $1 p$

Since $B E=B C=1 m$ and $E C=\sqrt{2} m$, then $\mathrm{BE} \perp \mathrm{BC}$ (2)

From (1) and (2) it follows that CDFE is a rectangle, so there exists $\mathcal{C}(O, r)$, such that $C F \cap D E=\{O\}$ and $r=\frac{1}{2} D E$. (3)...............1p We calculate $\mathrm{DE}=\sqrt{3}$ and from (3), it follows that $r=\frac{\sqrt{3}}{2} m \ldots \ldots \ldots . .1 \mathbf{p}$

Let $\mathrm{HI} \perp \mathrm{CD}$ and $\mathrm{OI} \cap \mathrm{CD}=\{\mathrm{M}\}$ and since $\mathrm{H}, \mathrm{I} \in \mathcal{C}(\mathrm{O}, r)$, then $\mathrm{H} \in(\mathrm{MO}$,

![](https://cdn.mathpix.com/cropped/2024_06_07_2aeee1df30a437c1f6afg-2.jpg?height=643&width=742&top_left_y=1429&top_left_x=1205)
such that $H I=D E=\sqrt{3} m$ and $O H=O I=\frac{\sqrt{3}}{2} m$, and

$C M=M D=\frac{1}{2} m$

$1 p$

Let $H I \cap E F=\{P\}$ and since $H I \perp C D$, CDFE is a rectangle and (4), then $E P=P F=\frac{1}{2} m$.

Let G, N, and Q be the projections of points $\mathrm{H}, \mathrm{P}$, and $\mathrm{I}$ respectively on the plane (ABC).

$.1 p$

It is demonstrated that $\mathrm{N} \in(\mathrm{GM})$ and $\mathrm{Q} \in(\mathrm{NM})$, such that on the perpendicular bisector $\mathrm{MN}$ of segment [CD] we can determine $N M=1 m$ and $G N=M Q=\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{2}} m$.

The nature of the polygon is justified and the required area is calculated, i.e., $A_{A G B C Q D}=\frac{2+\sqrt{6}}{4} m^{2}$.