8
$\begingroup$

Here is a diagram and challenge description that should be clear and simple to understand.

enter image description here

$\endgroup$
7
  • $\begingroup$ How did you find the midpoint without a compass? $\endgroup$
    – KSmarts
    Feb 23, 2015 at 18:06
  • $\begingroup$ The midpoints are provided. $\endgroup$
    – Moti
    Feb 23, 2015 at 20:06
  • $\begingroup$ I know that, I'm asking how you got them, and why can't we have the same tools? $\endgroup$
    – KSmarts
    Feb 23, 2015 at 20:07
  • $\begingroup$ So the ruler can only be used to draw lines between two points and the intersection of any two lines can be viewed as a point? $\endgroup$
    – fibonatic
    Feb 23, 2015 at 20:35
  • $\begingroup$ It's easy if you use a compass - just use shear transformations. $\endgroup$
    – user88
    Feb 23, 2015 at 22:08

2 Answers 2

3
$\begingroup$

Make a line segment starting at $A$, through $D$.
Make a line segment starting at $C$, through $B$.
Call the point where they intersect $G$.

enter image description here

The triangle $EFG$ has the same area as the quadrilateral.

Proof: Extend $ED$ and draw a line segment starting at $B$ and running parallel to $CD$, call the point where they intersect $X$.

enter image description here

I know that triangle $EFX$ has the same area as the quadrilateral because triangle $DFB$ and triangle $DFX$ have the same area (same base and height). Unfortunately, I can't draw line $BX$, but I can use the same proof by splitting the quadrilateral into 2 quadrilaterals.

Point $E$ bisects segment $AB$. Therefore, if I draw a line parallel to $AD$ starting at $E$, it will bisect $DB$.
Similarly, point $F$ bisects segment $CD$. Therefore, if I draw a line parallel to $CB$ starting at $F$, it will bisect $BD$.
Call that point $H$ and draw a line starting at point $G$, through $H$ until it intersects $EF$. Call that point $I$.

enter image description here

$DG$ is parallel to $EH$ and therefore $EIG$ has the same area as $IEDH$.
$BG$ is parallel to $FH$ and therefore $IFG$ has the same area as $IFBH$.
Together this means that $EFG$ has the same area as $EFBD$.

The lines drawn in the proof can't be drawn using just a ruler, but that is unimportant.

$\endgroup$
4
  • $\begingroup$ How do triangles DFB and DFX have the same height? $\endgroup$
    – Marmy1954
    Feb 25, 2015 at 2:48
  • $\begingroup$ Height of both is the distance between the parallel lines. $\endgroup$ Feb 25, 2015 at 2:53
  • $\begingroup$ And how do you construct BX to be parallel with only an unmarked ruler? $\endgroup$
    – Marmy1954
    Feb 25, 2015 at 3:07
  • $\begingroup$ You can't. You only need to get point G with the ruler. Everything after that is the proof. It doesn't need to be drawn. $\endgroup$ Feb 25, 2015 at 3:10
1
$\begingroup$

If I am allowed to measure distances with the ruler, then it would work to

measure the distance from the crossing point if the two long lines to the point B and replicate that distance from E towards A, calling the new point G. Since FDG has the sum of the perpendicular heights of the two triangles that have FD as a base, it has the same area as the quadrilateral.

Of course if you mean ruler as in the classical ruler and compass construction, without the compass, I'm more stuck!

$\endgroup$
2
  • $\begingroup$ The idea is of a ruler without marking. No measurement is possible, even not adding markings:) $\endgroup$
    – Moti
    Feb 18, 2015 at 5:20
  • $\begingroup$ @Moti Hmmmm. back to the drawing board. $\endgroup$
    – not my job
    Feb 18, 2015 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.