A coordinate system was drawn on the board and points $A (1,2)$ and $B (3,1)$ were marked. The coordinate system was erased. Restore it by the two marked points.
Problem
Source: 2008 Oral Moscow Geometry Olympiad grades 8-9 p1
Tags: analytic geometry, geometry, construction
eduD_looC
13.10.2020 20:48
What do you mean by "restore"?
somebodyyouusedtoknow
13.10.2020 20:50
I guess you must be able to provide an algorithm to construct the lattice points of the coordinate system back. cmiiw
eduD_looC
13.10.2020 20:52
This question is a little unclear. How are we supposed to answer this?
parmenides51
13.10.2020 20:54
if you find the point (0,0) and any point on one of the axes, you may construct both axes. It is a geometry construction problem so only compass and unmarked ruler are available tools. You also need to construct at least a unit point (0,1) or (1,0) on one of the axes, in order to may find any coordinates.
eduD_looC
13.10.2020 20:56
Okay. Lemme think.
somebodyyouusedtoknow
31.12.2020 01:18
Quite a tough problem for an eighth grader, not to mention that it is still problem number 1. Edit: I have thought of ways to optimise the steps.
Construct a circle centred at $A$ with radius $AB$. Letting $O$ = (0, 0), it's clear that $OA^2 = AB^2 = 5$ and $OB^2 = 10$ which means $OA \perp AB$. Hence the origin point is defined.
Rotate point $B$ by $90^{\circ}$ to $D = (1, -3)$ which means $AD$ is parallel to the Y-axis, following the linear equation $x = 1$. Then, construct a parallel to line $AD$ through $O$ so the Y-axis is defined.
Now construct a perpendicular to line $AD$ through $O$ so that the X-axis is defined. The length of this line between $x = 1$ and $x = 0$ is exactly 1 unit, so the unit is defined, hence the coordinate system is restored by constructing parallel lines to the X and Y-axis with a distance of 1 between each parallel line.
Construct $C(4, 3)$ by using a compass centred at $B$ with radius $AB$, notice $CB \perp BA$ so this is constructable.
Next, take the $AC$ as the radius, notice $AC = \sqrt{10}$ and construct a circle $\omega_1$ with that radius centred at $B$, so that the circle passes through (0, 0).
Lastly, take radius $AB$ and construct a circle $\omega_2$ centred at $A$, this will intersect (0, 0) as well by the Pythagorean Theorem, so the centre of the coordinate system lies on the intersection of $\omega_1$ and $\omega_2$.
From here, take $C$ on (4, 3) and rotate by 90 degrees, centred at (0, 0) so that its image lies on $C' (3, -4)$. Here, construct $BC'$. Then make a perpendicular of $BC'$, call this line $k$ passing through $C$ to get line $y = -4$. Now make a parallel line, call this line $l$, to the constructed perpendicular line so that it passes through point $A$, hence we now have a line $y = 2$. Take the perpendicular to lines $k$ and $l$ passing through $P, R$ and $B$ on lines $k, l$ respectively and call this line $m$, so that we can construct a perpendicular bisector of $PR$, which we call as $n$ which is the line $y = -1$. Let line $m$ intersect $n$ at $Q$. So the perpendicular bisector of $QB$ will be the line $y = 0$.
From here, take the perpendicular on (0, 0) to line $y = 0$ to get the X-axis. Hence we can now draw a parallel line $x = 1$ passing through $A$ so the unit is defined, and the coordinate system is hence restored.
P.S. The unit is already defined which is the distance between $Q$ with the midpoint of $QB$. This solution is not optimal, because it's my first attempt on the problem.
djmathman
31.12.2020 01:32
The problem proceeds in $2+\varepsilon$ steps (with intermediate constructions omitted, of course).
Rotate $B$ ninety degrees about $A$ to point $O(0,0)$.
Construct $M(\tfrac12,1)$, the midpoint of $\overline{OA}$.
Then the line through $O$ parallel to $MB$ is the $x$-axis, while the line through $O$ perpendicular to $MB$ is the $y$-axis.
natmath
31.12.2020 01:38
Not sure if this is correct (seemed a little too easy) Assuming the coordinate system is orthogonal and the cross product of $<1,0>$ and $<0,1>$ is out of the page, then the origin can be constructed by drawing the line $l$ that passes through $A$ and is perpendicular to $AB$, and drawing the circle $c$ with center $A$ and radius $AB$. The first intersection of $c$ and $l$ that is in the clockwise direction of $B$ on circle $c$ will be the origin $O$. Now we can use vector addition/subtraction to construct $(0,5)=3\overrightarrow{OA}-\overrightarrow{OB}$ Construct the y-axis by connecting $(0,5)$ and $O$. Reflect $A$ about the $y$-axis to construct the vector $<-1,2>$. Now we can construct $(0,4)=\overrightarrow{OA}+<-1,2>$. We can construct the length of $1$ as the distance between $(0,4)$ and $(0,5)$, and from there construct $(0,1)$.
somebodyyouusedtoknow
31.12.2020 02:21
@above Using compasses and a straight edge, you can do the following basic constructions: 1. Add, subtract, multiply, and divide lengths of segments 2. Construct midpoints of segments and perpendicular lines (perpendicular bisector) 3. Construct parallel lines 4. Construct circles 5. Bisect an angle So @djmathman's solution seems fine to me.
natmath
31.12.2020 02:27
@above is there anything wrong with my solution. You can do vector addition and subtraction by copying the angle and length