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What is the area of $HIJK$ quadrilateral, if the area of $ABC$ triangle is $70$, $BE=ED=DA$, and $BF=FG= GC$?

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closed as off-topic by Crostul, verret, Alexander Gruber Jan 8 at 22:04

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  • $\begingroup$ I got $\frac{50}{3}.$ $\endgroup$ – Michael Rozenberg Jan 8 at 19:59
  • $\begingroup$ There should be a solution using Ceva's thm or sth. $\endgroup$ – user614671 Jan 8 at 20:02
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    $\begingroup$ The area is $9$. Tthe points $B,I,K$ are collinear and the line containing them intersect $AC$ at its midpoint $M$. $\triangle HIK$ is bounded by 3 cevians $AF$, $BM$ and $CD$ with $$(x,y,z) \stackrel{def}{=}\left(\frac{CF}{BF}, \frac{AM}{MC}, \frac{BD}{AD}\right) = (2,1,2)$$ By Routh's theorem, area of $\triangle HIK$ is $$\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}\cdot 70 = \frac{9}{140}\cdot 70 = \frac92$$ By a similar argument, area of $\triangle IJK$ is also $\frac92$. $\endgroup$ – achille hui Jan 9 at 14:28
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Let's put the area of 70 on hold for a moment. Furthermore, given the listed constraints and the implicit assumption that the problem is solvable using only those constraints, we choose any proportions for the outer triangle we like.

Let's examine the case of a right isosceles triangle with hypotenuse $BC$ and a base length of 3. It has an area of $\frac{9}{2}$.

Letting the origin $(0,0)$ rest at $A$

  • $B$ is at $(0,3)$.
  • $C$ is at $(3,0)$.
  • $D$ is at $(0,1)$.
  • $E$ is at $(0,2)$.
  • $F$ is at $(1,2)$.
  • $G$ is at $(2,1)$.

This gives us the following lines

  • $AF$ is $y=2x$
  • $AG$ is $y=\frac{1}{2}x$
  • $DC$ is $y=-\frac{1}{3}x+1$
  • $EC$ is $y=-\frac{2}{3}x+2$

Considering the four batches of two equations with two unknowns yields the coordinates of the remaining four points.

  • $H$, from $y=2x,\;y=-\frac{1}{3}x+1$ is $(\frac{3}{7},\frac{6}{7})$.
  • $I$, from $y=2x,\;y=-\frac{2}{3}x+2$ is $(\frac{3}{4},\frac{3}{2})$.
  • $J$, from $y=\frac{1}{2}x,\;y=-\frac{2}{3}x+2$ is $(\frac{12}{7},\frac{6}{7})$.
  • $K$, from $y=\frac{1}{2}x,\;y=-\frac{1}{3}x+1$ is $(\frac{6}{5},\frac{3}{5})$.

We can express the general form for the area of a quadrilateral based on it's vertices using vector arithmetic $$A=\frac{1}{2}\vert(\vec{J}-\vec{H})\times(\vec{K}-\vec{I})\vert$$ We can plug this into Wolfram Alpha as

abs(Cross[([12/7,6/7]-[3/7,6/7]),([6/5,3/5]-[3/4,3/2])])*.5

Scaling that for our area 70 triangle, gives a final area of 9.

This differs from Peter Foreman's answer, so one of us must have made a typo or a false assumption. Frankly mine just looks wrong, judging from your diagram. In any case he deserves credit for advocating such a brute-force approach.

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  • $\begingroup$ Thanks! it seems I needed to be more specific to wolfram about what I was trying to do. $\endgroup$ – ShapeOfMatter Jan 9 at 15:04
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The area is $\frac{741}{68}$ I'm quite sure. You can find the coordinates of $H,I,J,K$ by using the equations of each straight line and taking $A$ as the origin. I then used Wolfram: Alpha to find the area given the coordinates of each vertex.

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