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$\mathit{ABCD}$ is a parallelogram that is not a rectangle. $E$ is a point on $\overline{\mathit{BC}}$, and $F$ is a point on $\overline{\mathit{CD}}$. $\triangle\mathit{ADE}$ and $\triangle\mathit{ABF}$ are inscribed in the parallelogram, and they enclose four triangular regions and one quadrilateral. I would to choose $E$ and $F$ with the following conditions:

i.) the triangular region with an edge along $\overline{\mathit{AD}}$ and outside $\triangle\mathit{ABF}$ is to have an area of 48, and the triangular region with a vertex at $E$ and outside $\triangle\mathit{ABF}$ is to have an area of 21;

ii.) the triangular region with an edge along $\overline{\mathit{AB}}$ and outside $\triangle\mathit{ADE}$ is to have an area of 63, and the triangular region with a vertex at $F$ and outside $\triangle\mathit{ADE}$ is to have an area of 6;

iii.) the quadrilateral circumscribed by both triangles is to have an area bigger than 85 and less than 105.

What is an instance of such a parallelogram and points $E$ on $\overline{\mathit{AB}}$ and $F$ on $\overline{\mathit{CD}}$?

Adelyn
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1 Answers1

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It is possible, the area of the inner quadrilateral will be $99$.

One way to realize this is start from a rectangle $ABC'D'$ with base $AB = 21$ and height $C'D' = 16$ and then translate $C'$, $D'$ along a line parallel to $AB$. Take $E$ to the mid point of $BC$ and $F$ the on-third point on $DC$.

The figure below illustrate one possible configuration by translating $C',D'$ for a distance $4$.
The important points are located at

$$\begin{align} &A = (0,0),\quad B = (21,0),\quad C = (25,16),\quad D = (4,16),\\ &E = \frac{B+C}{2} =(23,8),\quad F = \frac{2D+C}{3} = (11,16)\\ &P = \left(\frac{69}{4},6\right),\quad Q = \left(\frac{66}{7}, \frac{96}{7}\right),\quad R = \left(\frac{27}{2},12\right) \end{align} $$

Two triangles in an quadrilateral

How do I get this answer?

What I have done is parameterize $E$ as $u B + (1-u) C$, $F$ as $v D + (1-v) C$. Setup a bunch of equations in $u,v$ and ask an CAS to do the dirty work. The actual algebra is really horrible, I won't repeat it here.

achille hui
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