How to construct a square ABCD given point C, circle and a line so that point A lies on the line and point D lies on the circle?
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$– José Carlos SantosFeb 26, 2018 at 19:20
-
2$\begingroup$ Looks like $D$ is on the circle, not $B$. Anyway, what are your thoughts on the problem? I guess "use a computer" is not the expected answer? $\endgroup$– Matthew LeingangFeb 26, 2018 at 19:20
-
$\begingroup$ Are you given the centre of the circle like in the picture or just the circle? $\endgroup$– Arnaud MortierFeb 26, 2018 at 19:41
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Suppose that the problem is solvable.
Rotate a line around $C$ for $45^{\circ}$ and then dilate it with factor ${1\over \sqrt{2}}$. Intersection point between this new line and a circle is a point $D$ (on picture)...
the square construction Blue(solid) is what is given; the green is the solution.
- $CL$ is the perpendicular to the line;
- the blue dotted line is the original line $45^{\circ}$ rotated;
- $CN$ is ${1\over \sqrt{2}}$-times $CM$;
- $LN$ intersecting the circle gives us point $D$;
- $CD$ is the first side of the square.
How to perform this spiral similarity in practice? If point $X$ is on a line and let $XZCY$ is a (positive oriented) square with diagonal $XC$, then $X$ maps to $Y$. So take two pointy $X_1$ and $X_2$ on a line and then $Y_1Y_2$ is this new line.
