# How do you prove $S_{XYZT} \leq \dfrac{1}{5} S_{ABCD}$?

In the given figure, $$ABCD$$ is a convex quadrilateral. Suppose that $$M, N, P, Q$$ are mid-points of $$AB, BC, CD, DA$$, respectively. Prove that $$S_{XYZT} \leq \dfrac{1}{5} S_{ABCD}$$ where $$S_{ABCD}$$ (resp. $$S_{XYZT}$$) is the area of $$ABCD$$ (resp. $$XYZT$$)? Could you please give a key hint to solve this exercise? Thank you so much for your discussions!

• What have you tried? This can be coordinate-germ brute forced out, so where are you stuck? Apr 23, 2020 at 21:19
• According to geogebra the ratio is not exactly 5. Especially the ratio tends to 6 if one of the sides tends to 0.
– user
Apr 23, 2020 at 21:45
• @CalvinLin I tried to do with the coodinates of points and calculate areas. However, i prefer to find another nice solution. Thank you so much for your interests. Apr 24, 2020 at 4:44
• @user Thank you so much for your nice answers. Can you give a reference for the case that ratio which tends to 6? Apr 24, 2020 at 4:50
• Related (duplicate?): "Quadrilateral formed by connecting the vertices of a convex quadrilateral to midpoints of non-adjacent sides", with a reference to the 2011 Mathematics Magazine article "Crosscut Convex Quadrilaterals". This answer shows that the ratio of inner area to total area is between $1:6$ and $1:5$.
– Blue
Apr 24, 2020 at 12:23

As already pointed out in a comment the ratio $$\dfrac{S_{ABCD}}{S_{TXYZ}}$$ is not exactly equal to 5, though it is surprisingly close to the value almost in any convex quadrilateral. Only if one side tends to 0 (so that the quadrilateral degenerates to triangle) the ratio tends to 6 (as it also should).

There is however a class of quadrilaterals for which the ratio is exactly 5. This class is parallelograms, and the proof in this case is simple. As easy to understand for any convex quadrilateral: $$S_{TXYZ}=S_{AXM}+S_{BYN}+S_{CZP}+S_{DTQ}.$$ and $$S_{ABCD}-S_{TXYZ}=S_{AYB}+S_{BZC}+S_{CTD}+S_{DXA}.$$ Specifically for parallelogram we have: $$S_{AYB}=4S_{AXM}, \dots$$ Thus, $$S_{ABCD}-S_{TXYZ}=4S_{TXYZ}.$$

UPDATE:

On the basis of numerical evidence I would conjecture the following statement:

For any convex quadrilateral $$5\le\dfrac{S_{ABCD}}{S_{TXYZ}}<6$$ and the ratio is equal to 5 if and only if the quadrilateral $$TXYZ$$ is a trapezoid.

For the characterization of the quadrilateral $$ABCD$$ the above statement means that its vertices lie on four equidistant parallel lines, two opposite vertices being on the external lines (see figure below). I do not know if a special name for such a quadrilateral exists.

To prove the "if" part of the statement only a slight modification of the previous proof (for parallelogram) is required due to the fact that $$S_{AXM}=S_{DTQ}$$ and $$S_{BYN}=S_{CZP}$$. • Yes, your solution is the best for your special class. However, Can you say more about this point? There is however a class of quadrilaterals for which the ratio is exactly 5. This class is parallelograms''. Are there the other such classes or not? Apr 24, 2020 at 5:02
• @mathJuan See update of my answer.
– user
Apr 24, 2020 at 12:06
• Thank you so much for your nice update! Apr 24, 2020 at 15:26

►Let the four vertex $$C=(0,0),D=(d_1,d_2),A=(a_1,a_2),B=(b_1,b_2)$$.

►Calculation determines lines $$\overline{BQ},\overline{ND},\overline{MC},\overline{AP}$$.

►Points $$Y=\overline{BQ}\cap\overline{MC}\\X=\overline{BQ}\cap\overline{AP}\\Z=\overline{ND}\cap\overline{MC}\\T=\overline{ND}\cap\overline{AP}$$

►Do you know how to calculate directly the area of a convex quadrilateral? For example for $$CDAB$$ put the coordinates as follows starting from an arbitrary vertice and contrary to clockwise direction, say starting with $$C=(0,0)$$

$$0\hspace{10mm}0 \\d_1\hspace{10mm}d_2\\a_1\hspace{10mm}a_2\\b_1\hspace{10mm}b_2\\0\hspace{10mm}0$$ You must finish repeating the first chosen vertice.Then here you have the area is given for $$\frac12[(0\cdot d_2+d_1\cdot a_2+a_1\cdot b_2+b_1\cdot0)-(0\cdot b_2+b_1\cdot a_2+a_1\cdot d_2+d_1\cdot0)]$$ (Multiplication descending for positive parenthesis and ascending for the negative one).

Repeat this with the smaller quadrilateral and compare.

• Yes, thank you so much for your solution. In fact, one can suppose that $D(d_1; 0)$. I prefer to see the other ways to solve this problem. However, I like your solution. Apr 24, 2020 at 4:56
• You are welcome. Note that with this answer you have a direct way to verify if in fact your formula with the $5$ is true or not. I wanted to leave you this task. Apr 24, 2020 at 12:16
• Yes, you are right. I have just updated again my question. It should be $S_{XYZT} \leq \dfrac{1}{5} S_{ABCD}$. Thank you very much one again for your nice hints. Apr 24, 2020 at 15:28