Coordinate axes (without any marks, with the same scale) and the graph of a quadratic trinomial $y = x^2 + ax + b$ are drawn in the plane. The numbers $a$ and $b$ are not known. How to draw a unit segment using only ruler and compass?
Problem
Source: Tuymaada 2020 Senior, P5 (Junior, P5)
Tags: conics, parabola, construction, geometry, algebra, analytic geometry, quadratics
06.10.2020 05:04
06.10.2020 05:41
06.10.2020 06:19
@above Can you tell me how do you divide a segment by any integer amount using ruler and compass. It doesn't seem obvious to me.
06.10.2020 07:05
GorgonMathDota wrote: @above Can you tell me how do you divide a segment by any integer amount using ruler and compass. It doesn't seem obvious to me. Suppose you want to divide a segment of length $x$. Let this segment be $AB$. Draw a segment $MN$ of length $xn$ parallel to $AB$ and mark point $P$ on the segment such that $MP=x$. Intersect lines $AM$ and $BN$ at point $X$. Finally, let line $PX$ intersect $AB$ at $Q$. Line $PX$ would divide segments $AB$ and $MN$ into the same ratios, so $AQ=\frac xn$. [asy][asy]pair A=(0,0);pair B=(1,0);pair M=(-.5,-1);pair N=(2.5,-1);pair P=(.5,-1);pair X=extension(A,M,B,N);pair Q=extension(A,B,X,P);draw(A--B^^M--N);draw(M--X--N^^X--P,dotted);dot(A^^B^^P^^M^^N^^X^^Q);label("$A$",A,W);label("$B$",B,NE);label("$M$",M,SW);label("$N$",N,SE);label("$P$",P,S);label("$X$",X,(0,1));label("$Q$",Q,SE);[/asy][/asy]
06.10.2020 12:47
Clearly, we don’t need coordinate axes since it’s possible to construct it (up to a translation).
06.10.2020 12:52
Rickyminer wrote: Clearly, we don’t need coordinate axes since it’s possible to construct it (up to a translation). In the solutions above, the coordinate axes are required to construct lines that are parallel to the axis of symmetry. But, I would be really interested to see a solution where coordinate axes aren't necessary and only uses the parabola itself.
06.10.2020 13:01
First, we can assume that $a = b = 0$, since we can just shift the origin to the vertex of the parabola $y = x^2+ax+b$. We can do this by: 1. Pick a random point $P$ on the parabola 2. Draw a line passing through $P$ parallel to the $X$ axis and let the line intersects the parabola again at $Q$. 3. Draw the perpendicular bisector of $PQ$ and let it intersects the parabola at $O'$. Then $O'$ is the vertex of the parabola. 4. Draw a line passing through $O'$ parallel to the $X$ axis and the $Y$ axis and we get a new coordinate axes. Draw line $y = x$ (we can do this by drawing the angle bisector of the first quadrant of the coordinate axes). Clearly, the intersection of this line and the parabola $y = x^2$ is $(1,1)$ and it's enough to drop a perpendicular from $(1,1)$ to the coordinate axes and draw the desired unit segment.
06.10.2020 13:30
Construct two parallel chords of the parabola. It can be seen that the line joining their midpoints is parallel to the axis of symmetry.
26.02.2021 17:20
Nice problem In my solution, the direction of the coordinate axes aren't used, rather, only the fact that the coefficient of the $x^2$ term is $1$ is used. My solution is complicated, but it is very natural to arrive at, if you know about the Artzt Parabola. The Pascal's theorem can be used to give a ruler-only construction of the tangent to a conic at a given point. Take any two points $B,C$ on the parabola and draw tangents to the parabola at these points. Let the tangents intersect at $A$. Artzt Parabola of $\triangle ABC$ : The parabola that passes through $B$ and $C$ and is tangent to the sides $AB$ and $AC$ is the $A$-Artzt Parabola of the triangle $ABC$. Its focus if the mid point of the $A$ symmedian chord of $\triangle ABC$. Thus, in the given graph, we can find the focus of the parabola. Also, the axis of this parabola is parallel to the $A$ median. Since we know the focus and the direction of the axis, we can draw the axis of this parabola. Let the focus be $S$ and the vertex be $V$, then, $SV=\frac{1}{4}$. Thus, we can draw a segment of length equal to four times the length of $SV$ to get the desired segment.
12.05.2022 17:13
Let $O$ be the origin and assume the parabola intersects like $y$ axis at $S$ and the $x$ axis at $T,U$. One can easily see that $OS=OT*OU$ thanks to Vieta, so the circle through $S,T,U$ passes through the point $P=(0,1)$ by PoP. We can construct $P$ and so we get the unit segment $OP$.