# Area of the intersection of two triangles.

Let $$\triangle{ABC}$$ be a triangle with $$AB=5$$, $$BC=7$$, and $$CA=4$$. Define $$D$$, $$E$$, and $$F$$, to be the midpoints of $$AB$$, $$BC$$, and $$CA$$ respectively. Let $$G$$ the intersection of the medians of $$\triangle{ABC}$$, and let $$H$$, $$I$$, and $$J$$ be the midpoints of $$AG$$, $$BG$$, and $$CG$$ respectively. Find the area of the hexagonal region common to both $$\triangle{DEF}$$ and $$\triangle{HIJ}$$.

This is too tough for me to even start. I suppose you would start with similar triangles? I know for a fact that the areas of both $$\triangle{DEF}$$ and $$\triangle{HIJ}$$ are both a quarter the area of $$\triangle{ABC}$$, but I have no idea how to proceed. Can somebody help?

• The areas of $\triangle DEF$ and $\triangle HIJ$ are actually a quarter of that of $\triangle ABC$. Indeed the ratios does not depend on the shape in this question, try to draw a picture for the case when $\triangle ABC$ is equilateral and see if you can get some intuition. – Hw Chu Mar 14 at 21:42

We can start by finding the length of the median, using Apollonius' Theorem, which yields $$AB^2 + AC^2 = 2(AE^2 + BD^2) \rightarrow 25+16 = 2(AE^2 + 12.25) \rightarrow AE = \frac{\sqrt{33}}{2}$$. It is easy to see that the ratio of $$AG : GE = 2:1$$ , and since $$H$$ is the midpoint of $$AG$$, we know that $$AE$$ is trisected by $$H, G$$. Repeat this for the other medians, and then it should be easy to go from there.

HINT

► Let $$C=(0,0),B=(7,0),A=(x,y)$$ so the coordinates $$(x,y)=\left(\dfrac{20}{7},\dfrac{8\sqrt6}{7}\right)$$ (because $$x^2+y^2=16$$ and $$(x-7)^2+y^2=25$$).

► Formulas for the medians in function of sides give $$(AA')^2=\dfrac14(2x16+2x25-49)=\dfrac{33}{4}$$ so $$AA'=\dfrac{\sqrt{33}}{2}$$. Similarly $$BB'=\sqrt{33}$$ and $$CC'=\dfrac{\sqrt{105}}{2}$$.

►Coordinates of $$H=(h_1,h_2)$$ are given by the system $$\dfrac{20-7h_2}{8\sqrt6-7h_1}$$ and $$AH=\dfrac{AA'}{3}$$. Similarly for $$I$$ and $$J$$.

$$HI\cap{ED}$$ and $$HI\cap{FD}$$ give vertices $$P_1$$ and $$P_2$$ of the hexagon respectively.

$$\space \space HJ\cap{FD}$$ and $$HJ\cap{FE}$$ give vertices $$P_3$$ and $$P_4$$.

$$\space \space JI\cap{FE}$$ and $$JI\cap{ED}$$ give vertices $$P_5$$ and $$P_6$$.

►Now apply formulas for area of triangles (four times) using coordinates of points $$P_i; i=1,2,3,4,5,6$$

The hint.

Let $$AE\cap IJ=\{T\}$$, $$IJ\cap ED=\{M\}$$ and $$EF\cap IJ=\{P\}$$.

Thus, $$T$$ a midpoint of $$EG$$, which says $$ET=\frac{1}{2}\cdot\frac{1}{3}AE=\frac{1}{6}AE.$$ Also, since $$ADEF$$ is a parallelogram and $$IJ||DF,$$ we obtain $$ET$$ is a median of $$\Delta MEP.$$

But $$\Delta MEP\sim\Delta CAB,$$ which gives $$S_{\Delta MEP}=\frac{1}{36}S_{\Delta ABC}.$$ Id est, the needed area is equal to $$S_{\Delta EFD}-3S_{\Delta MEP}=\frac{1}{4}S_{\Delta ABC}-3\cdot\frac{1}{36}S_{\Delta ABC}=\frac{1}{6}\sqrt{8\cdot1\cdot4\cdot3}=\sqrt{\frac{8}{3}}.$$

• To down-voter. It was typo. I fixed. See now. – Michael Rozenberg Mar 15 at 16:44