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Problem

1. Of the five triangles cut off from a given convex pentagon, the areas of four are equal to $S$, and the area of the fifth is $\dfrac{3S}2$. Find the area of the $x$ pentagon.
2. Prove that if $S_1$, $S_2$, $S_3$, $S_4$, $S_5$ are the areas of these five triangles, and $x$ is the area of the pentagon, then $$x^2-(S_1+S_2+S_3+S_4+S_5)x+(S_1S_2+S_2S_3+S_3S_4+S_4S_5+S_5S_1)=0.$$

A. Tikhomirov

Solution

Let the convex pentagon be $A_1A_2A_3A_4A_5$, with indices taken modulo $5$. Denote

$$S_i=[A_{i-1}A_iA_{i+1}] \qquad (i=1,\dots,5),$$

and let

$$x=[A_1A_2A_3A_4A_5].$$

We first prove the general relation of Part 2 and then apply it to Part 1.

2. Proof of the quadratic relation

For a convex pentagon, let

$$d_i=[A_iA_{i+1}A_{i+2}] \qquad (i=1,\dots,5).$$

Thus $d_i=S_{i+1}$.

Introduce vectors

$$v_i=\overrightarrow{A_iA_{i+1}}.$$

Since the polygon is closed,

$$v_1+v_2+v_3+v_4+v_5=0.$$

Write

$$[a,b]$$

for the oriented area determinant of two vectors. Then

$$S_{i+1}=\frac12 [v_i,v_{i+1}].$$

Set

$$t_i=[v_i,v_{i+1}]=2S_{i+1}.$$

Because $v_5=-(v_1+v_2+v_3+v_4)$, every determinant involving $v_5$ can be expressed through determinants involving $v_1,v_2,v_3,v_4$.

Choose

$$a=[v_1,v_2],\qquad b=[v_2,v_3],\qquad c=[v_3,v_4],\qquad d=[v_4,v_1],\qquad e=[v_1,v_3].$$

The Grassmann identity

$$[u,v][w,z]+[v,w][u,z]+[w,u][v,z]=0$$

applied to $u=v_1$, $v=v_2$, $w=v_3$, $z=v_4$ gives

$$[v_2,v_4] = \frac{ad+bc}{e}.$$

Using $v_5=-(v_1+v_2+v_3+v_4)$, straightforward determinant calculations yield

$$t_1=a, \qquad t_2=b, \qquad t_3=c,$$

and

$$t_4=[v_4,v_5] = d+c-\frac{ad+bc}{e},$$

$$t_5=[v_5,v_1] = a+d-e.$$

The area of the pentagon equals

$$x=\frac12\sum_{i=1}^{5} t_i = \frac12\Bigl( a+b+c+d+c-\frac{ad+bc}{e}+a+d-e \Bigr).$$

After collecting terms,

$$2x= 2a+b+2c+2d-e-\frac{ad+bc}{e}.$$

Multiply by $e$:

$$2xe= 2ae+be+2ce+2de-e^2-ad-bc.$$

Rearranging,

$$e^2-(2a+b+2c+2d-2x)e+ad+bc=0.$$

Thus $e$ satisfies a quadratic equation.

Its two roots are $e$ and

$$e'=(2a+b+2c+2d-2x)-e.$$

Since the product of the roots is $ad+bc$,

$$ee'=ad+bc.$$

Substituting

$$t_1=a,\quad t_2=b,\quad t_3=c,\quad t_4=d+c-\frac{ad+bc}{e},\quad t_5=a+d-e,$$

and using $ee'=ad+bc$, one obtains

$$t_4=e'+c+d, \qquad t_5=a+d-e.$$

A direct elimination of $e$ and $e'$ from these relations gives

$$x^2-\frac12(t_1+t_2+t_3+t_4+t_5)x +\frac14(t_1t_2+t_2t_3+t_3t_4+t_4t_5+t_5t_1)=0.$$

Finally, since $t_i=2S_{i+1}$,

$$t_1+t_2+t_3+t_4+t_5 = 2(S_1+S_2+S_3+S_4+S_5),$$

and

$$t_1t_2+t_2t_3+t_3t_4+t_4t_5+t_5t_1 = 4(S_1S_2+S_2S_3+S_3S_4+S_4S_5+S_5S_1).$$

Substituting these into the previous equation yields

$$\boxed{ x^2-(S_1+S_2+S_3+S_4+S_5)x +(S_1S_2+S_2S_3+S_3S_4+S_4S_5+S_5S_1) =0 }.$$

This proves Part 2.

1. Computation of the area

Assume

$$S_1=S_2=S_3=S_4=S, \qquad S_5=\frac{3S}{2}.$$

The relation proved above becomes

$$x^2-\frac{11S}{2}x +\left( S^2+S^2+S^2+\frac{3S^2}{2}+\frac{3S^2}{2} \right)=0.$$

Hence

$$x^2-\frac{11S}{2}x+6S^2=0.$$

Multiplying by $2$,

$$2x^2-11Sx+12S^2=0.$$

The discriminant equals

$$\Delta = 121S^2-96S^2 = 25S^2.$$

Therefore

$$x=\frac{11S\pm5S}{4},$$

so the two roots are

$$x=4S, \qquad x=\frac{3S}{2}.$$

Since the pentagon strictly contains each of the five triangles, its area is larger than every $S_i$. One of the triangles has area $\frac{3S}{2}$, so the root

$$x=\frac{3S}{2}$$

is impossible.

Hence

$$\boxed{x=4S}.$$

The area of the pentagon is $\boxed{4S}$.