Finding the value of a combined area in a square with center $O$

Question

I hope this message finds you well. I'm reaching out to ask for your help in solving a challenging geometry problem that I encountered in a recent exam. Despite my best efforts, I haven't been able to find a solution. I'm eager to gain insights that will improve my understanding of this geometric challenge.

Problem Description

Let $\mathbf{\large{O}}$ be the center of the square $ \mathbf{\large{ABCD}} $ . Find the value of $\mathbf{\large{\frac{[BOEG]+[\Delta DEF]+[\Delta CHF]}{[ABCD]}}}$ in the form $\mathbf{\large{\frac{a}{b}}}$, where $\mathbf{\large{a}}$ & $\mathbf{\large{b}}$ are natural numbers and coprime. Determine the value of $\mathbf{\large{a + b}}$.

Approach

As I predicted the area of $\Delta DEF$=$[GEFH]$ but I can't figure out a way of proving that if we can prove it somehow then maybe we can go ahead easily.

I believe that your expertise in geometry could provide valuable guidance in tackling this problem. Your insights and perspective would be greatly appreciated, and I am confident that your assistance will lead to a breakthrough in my understanding of this complex geometric concept.

Thank you in advance for taking the time to consider my request.

Answer 1

Hints.

Let $[ABCD]=1$.

1. $\triangle DFC\sim \triangle DBG$ $\Rightarrow$ $\dfrac{[DBG]}{[DFC]}=\dfrac{BD^2}{DC^2}=2$;

2. $\triangle DOE\sim \triangle DCH$ $\Rightarrow$ $\dfrac{[DCH]}{[DOE]}=\dfrac{DC^2}{DO^2}=2$ and $\dfrac{DH}{DE}=\dfrac{DC}{OD}=\sqrt2$;

3. $\triangle DEF\sim \triangle DHG$ $\Rightarrow$ $\dfrac{[DHG]}{[DEF]}=\dfrac{DH^2}{DE^2}=2$;

4. See the figure: $$ 1/4=S_1+S_2+S_3=(2S_1-S_2)+S_3+(2S_2-S_1). $$

$S_1=[DFC]$ and so on.

Answer 2

Locate point Z on AD such that $\triangle FDZ$ is isosceles with $FD = FZ$. The sizes of its internal angles are as marked. ZF will cut DE at X. Letting ZF produced to cut BC at Y, we get $\triangle FHY$ is also isosceles with $FH = FY$.

Angles marked with the same color are equal to the labelled value. [Explanations are skipped because those values can be found via simple angle chasing.]

---

Note that EFHG and also BEFY are cyclic.

Note also that DXFC is cyclic. This causes (1) $XD = XC$; and $XC = XG$ because $\triangle XGC$ is equilateral.

All of the above provide sufficient reason to demonstrate $\angle EHF = 30^0 = \angle ZFD$, and hence $EH // ZY$.

---

After letting HX to cut EF at T, we have $[\triangle THF] = [\triangle TEX]$.

Then, $[quad. EGHF] = [quad. EGHT] + [\triangle THF] = [quad. EGHT] + [\triangle TEX] = [\triangle HGX] = [\triangle HXD] = [\triangle THF] + [\triangle TFX] + [\triangle FXD] = [\triangle TEX] + [\triangle TFX] + [\triangle FXD] = [\triangle FED]$.

To make the proof easier to read, I label the affected regions as shown. Then, $[Quad. EGHF] = (1 + 2) + 5 = (1 + 2) + 3 = [\triangle HGX] = [\triangle HXD] = 5 + 4 + 6 = 3 + 4 + 6 = [\triangle FED]$

Therefore, total area of the shaded regions $= … = [\triangle BOC] = 0.25[square. ABCD]$.

Answer 3

You are right, the areas of triangle $DEF$ and quadrilateral $EFHG$ are equal, giving $area(DOC) = \frac14 area(ABCD)$.

Here is a solution using analytical geometry.

Let us work (see below) on an equivalent figure, symmetrical to the initial one with respect to line $AC$. I take on it a natural coordinate system where $D$ is the origin,

$$C=(1,0), \ \ A=(0,1), \ \ B=(1,1) \ \ \text{and} \ \ O=(\tfrac12,\tfrac12).$$

In this coordinate system,

• line $DH$ has equation $y=tx$ with $t:=\tan(\frac{\pi}{12})=2-\sqrt{3}$.

• ine $DG$ has equation $y=sx$ with $s:=\tan(\frac{\pi}{6})=\frac{\sqrt{3}}{3}$.

As $E$ is the intersection of straight lines $AC$ and $DG$ with resp. equations $x+y=1, \ \ y=sx$, we have

$$E=(\frac{1}{1+s},\frac{s}{1+s}).$$

For a similar reason :

$$F=(\frac{1}{1+t},\frac{t}{1+t})$$

Therefore, using for example shoelace formula, one obtains :

$$2 \times area(DFE)=\frac{s-t}{(s+1)(t+1)}$$

$$2 \times area(EFHG)=\tfrac12 \frac{s-1}{s+1}+1-s$$

Showing now that these expressions are equal is a straightforward task with the values of $s$ and $t$ given above.