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. $XYZT$)?

enter image description here

Could you please give a key hint to solve this exercise? Thank you so much for your discussions!

  • 1
    $\begingroup$ What have you tried? This can be coordinate-germ brute forced out, so where are you stuck? $\endgroup$
    – Calvin Lin
    Apr 23, 2020 at 21:19
  • 2
    $\begingroup$ According to geogebra the ratio is not exactly 5. Especially the ratio tends to 6 if one of the sides tends to 0. $\endgroup$
    – user
    Apr 23, 2020 at 21:45
  • $\begingroup$ @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. $\endgroup$ Apr 24, 2020 at 4:44
  • $\begingroup$ @user Thank you so much for your nice answers. Can you give a reference for the case that ratio which tends to 6? $\endgroup$ Apr 24, 2020 at 4:50
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    $\begingroup$ 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 $1:5$. $\endgroup$
    – Blue
    Apr 24, 2020 at 12:23

2 Answers 2


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. enter image description here

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, $$S_{ABCD}-S_{TXYZ}=4S_{TXYZ}.$$


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 $S_{BYN}=S_{CZP}$.

enter image description here

  • $\begingroup$ 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? $\endgroup$ Apr 24, 2020 at 5:02
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    $\begingroup$ @mathJuan See update of my answer. $\endgroup$
    – user
    Apr 24, 2020 at 12:06
  • $\begingroup$ Thank you so much for your nice update! $\endgroup$ 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.

  • $\begingroup$ 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. $\endgroup$ Apr 24, 2020 at 4:56
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    $\begingroup$ 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. $\endgroup$
    – Piquito
    Apr 24, 2020 at 12:16
  • $\begingroup$ 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. $\endgroup$ Apr 24, 2020 at 15:28

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