how can one draw the line $d$ into the rectangle $ABCD$ with only compassses and a ruler when $AY=XY=CX$ ($X$ is the intersection of $AB$ and $d$, $ Y$ is the intersection of $CD $ and $d$)

BTW: the ruler can't measure anything.

Here is the figure when drawn:

  • $\begingroup$ I believe the points $Y$ and $X$ will only exist if $AB>2AD$. How to construct them in that case is another matter. Argument: if we move $Y$ from $A$ to $B$ and $X$ from $C$ to $D$ at the same speed, $XY$ is initially larger than $AY=CX$, and is at its shortest for $AY=CX=AB/2$. If it's still longer than $AY=CX$ then, it'll never be shorter than $AY$ or equal to it. $\endgroup$ – Jack M Feb 11 '14 at 12:39

A hint:

The line will pass through the center $M$ of the rectangle, and one will have $|AY|=2|YM|$.

The locus of points whose distance from $A$ is twice as large as their distance from $M$ is an "Apollonian" circle $\gamma$ with center somewhere on the diagonal $AC$. By arguing about the two points where $\gamma$ intersects the diagonal one easily finds its center and radius.

Here is the picture; it shows that for the considered rectangle there are two solutions:

enter image description here

  • $\begingroup$ To be honest this was my initial solution. i drew it and it worked but i couldn't prove it so i left it. $\endgroup$ – PouyaH Feb 11 '14 at 18:03
  • $\begingroup$ My reason was that the center of this circle was the intersection of the bisector of AYM and the line AM in O which because of being a bisector creats two lines AO and MO with the ratio of 2($\frac{AY}{MY}=\frac{2}{1}$). now you can easily prove that CY is the bisector of BYM so therefore CYO is 90 degrees. now one must just draw a circle with the center of O and the radius of 2/3 of AM and wherever this circle intersects with AB, Y can be placed there. of course i couldn't prove this. i understand that you reached this through another way:$$(X-\frac{5}{3}AM)^2+Y^2=(\frac{4}{3}AM)^2$$ $\endgroup$ – PouyaH Feb 11 '14 at 18:28

Let $a:=\lvert A,B\rvert$, $b:=\lvert B,C\rvert$ and $x:=\lvert A,Y\rvert$. Then you have

\begin{align*} (2x-a)^2+b^2 &= x^2 \\ 4x^2-4ax+a^2+b^2 &= x^2 \\ 3x^2-4ax+a^2+b^2 &= 0 \end{align*} \begin{align*} x_{1,2} &= \frac{4a\pm\sqrt{4^2a^2-4\cdot3\cdot\left(a^2+b^2\right)}}{2\cdot 3} = \frac{2a\pm\sqrt{a^2-3b^2}}{3} \end{align*}

Now all you have to do is turn this formula into a ruler-and-compass construction. Here is one possible such construction:

Suggested construction

Start with the green part. Circles around $A$ and $D$ with radius $b$ construct $E$ such that $\triangle AED$ is regular, hence has height $\frac12\sqrt3\,b$. Therefore $\lvert D,F\rvert=\sqrt3\,b$.

The blue construction uses that length as $\lvert D,H\rvert$, which is one of the legs of a right triangle $\triangle DCH$. The hypothenuse is $a$. The point $G$ is the center of $CD$, its construction by ruler and compasses is not included in the figure. It serves as the center of the circumcircle of the right triangle, by Thales' theorem. The resulting second leg $CH$ has length $\sqrt{a^2-3b^2}$ by the Pythagorean theorem. The length $CK$ is the same, obtained by a circle around $C$.

The orange construction complete the process. $L$ was obtained from a circle around $D$ through $C$, so it satisfies $\lvert L,C\rvert=2a$ and therefore $\lvert L,K\rvert=2a-\sqrt{a^2-3b^2}$. $M$ was chosen an arbitrary distance above $K$, while $O$ is twice that distance below. I didn't include the construction of the vertical lines parallel to $BC$. The resulting $\lvert K,P\rvert=x$ is the needed distance. It can be used as the radius of a circle around $A$ and another around $C$, which defines $Y$ resp. $X$.

Note that I chose to subtract the square root instead of adding it, but the other solution would have been equally valid. Also note that this is not the shortest construction giving the desired result, far from it. If you work out the hint Christain wrote (which is a lot easier now that he expanded his answer), you can obtain a shorter construction like the one depicted below. The orange steps used above might still be useful, though, to do a division of a line into three parts of equal length.

Shorter construction

Here I first double the sort side using a circle around $D$ through $A$ to obtain point $E$. The line $EB$ intersects the diagonal $AC$ in point $F$, splitting it in a $2:1$ ratio. Up to here, this reflects the ideas used in the orange part above. A circle around $F$ through $C$ gives $Y$ (two choices) and $G$, a circle around $G$ through $A$ then gives $X$ (must be chosen to match $Y$).

The benefit of the first approach is that it exhibits some general ideas of how you can turn an algebraic result back into geometry. You can try this even if you didn't think of the ideas neccessary for the shorter construction.

  • $\begingroup$ thnx man this is awesome... $\endgroup$ – PouyaH Feb 11 '14 at 17:10

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