On a blank piece of paper, two points with distance $1$ is given. Prove that one can use (only) straightedge and compass to construct on this paper a straight line, and two points on it whose distance is $\sqrt{2021}$ such that, in the process of constructing it, the total number of circles or straight lines drawn is at most $10.$ Remark: Explicit steps of the construction should be given. Label the circles and straight lines in the order that they appear. Partial credit may be awarded depending on the total number of circles/lines.
Problem
Source: CMO 2022 P5
Tags: geometry
22.12.2021 15:50
Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..?
22.12.2021 20:30
FishHeadTail wrote: Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..? If so, then $5$ lines/circles suffice. On the line $l_1$ with the two points $A_0$ and $A_1$ ,copy the length $2021$ times in order to obtain the points $A_2,...,A_{2022}$ such that $A_iA_{i+1}=1$ for all $i=1,...,2021$. Now construct the circle $C_1$ of center $A_{2020}$ and radius $A_{2020}A_{2022}$ and the circle $C_2$ of center $A_{2022}$ and radius $A_{2020}A_{2022}$. By intersecting these two circles we get the points $P_1$ and $P_2$, we get two points lying on a new line $\ell_2$, which is perpendicular to $\ell_1$ and passes through $A_{2021}$. Now take the circle $C_3$ of center $A_{1011}$ and radius $A_0A_{1011}=1011$, which is in other words the circle with diameter $A_0A_{2022}$. Intersecting it with $\ell_2$ we get two points $Q_1$ and $Q_2$. Since $A_0Q_1A_{2022}$ is right-angled and $A_{2021}$ is its projection on $\ell_1$, it follows that $A_{2021}Q_1=\sqrt{A_0A_{2021}\cdot A_{2021}A_{2022}}=\sqrt{2021}$, for a total of $5$ things used. However if we can't copy distances as I did then I think it would be quite difficult, since you probably can create an integer in logarithmic time (for example $2^n$ in at least $n$ steps, I think).
22.12.2021 21:19
23.12.2021 04:40
23.12.2021 08:54
cadaeibf wrote: FishHeadTail wrote: Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..? If so, then $5$ lines/circles suffice. On the line $l_1$ with the two points $A_0$ and $A_1$ ,copy the length $2021$ times in order to obtain the points $A_2,...,A_{2022}$ such that $A_iA_{i+1}=1$ for all $i=1,...,2021$. Now construct the circle $C_1$ of center $A_{2020}$ and radius $A_{2020}A_{2022}$ and the circle $C_2$ of center $A_{2022}$ and radius $A_{2020}A_{2022}$. By intersecting these two circles we get the points $P_1$ and $P_2$, we get two points lying on a new line $\ell_2$, which is perpendicular to $\ell_1$ and passes through $A_{2021}$. Now take the circle $C_3$ of center $A_{1011}$ and radius $A_0A_{1011}=1011$, which is in other words the circle with diameter $A_0A_{2022}$. Intersecting it with $\ell_2$ we get two points $Q_1$ and $Q_2$. Since $A_0Q_1A_{2022}$ is right-angled and $A_{2021}$ is its projection on $\ell_1$, it follows that $A_{2021}Q_1=\sqrt{A_0A_{2021}\cdot A_{2021}A_{2022}}=\sqrt{2021}$, for a total of $5$ things used. However if we can't copy distances as I did then I think it would be quite difficult, since you probably can create an integer in logarithmic time (for example $2^n$ in at least $n$ steps, I think). Well but for “copying”, shall we count them as steps?
23.12.2021 10:20
No, because it doesn't create a new circle or line.
23.12.2021 10:30
CANBANKAN wrote: No, because it doesn't create a new circle or line. Instead I would say that we either count it as $0$ steps, because in practice you don't need to draw the whole circle or as $2021$ steps because technically you need to draw the entire circles to copy the distance, but surely not a single step.
23.12.2021 10:52
Or you can think of drawing a line cost one dollar, drawing a circle cost one dollar, nothing else costs money, and you have 10 dollars.
23.12.2021 19:51
I’m not sure but I think the intention was to only draw “a line through two points” and “a circle with a certain center and radius”. The ruler is assumed to have no markings on it (and, perhaps, the compass can’t be used to transfer distance). The main idea is basically divide & conquer. But it really takes time to reduce $11-20$ steps to $10$.
23.12.2021 20:20
There are many solutions that interpret this as number of operations instead of number of circles and lines drawn... congratulations for proving this stronger statement!
25.12.2021 07:45
FishHeadTail wrote: I’m not sure but I think the intention was to only draw “a line through two points” and “a circle with a certain center and radius”. The ruler is assumed to have no markings on it (and, perhaps, the compass can’t be used to transfer distance). The main idea is basically divide & conquer. But it really takes time to reduce $11-20$ steps to $10$.
The problem asks for a line and two points on it, so I think you need one more line $XP_3$.
25.12.2021 11:28
$l’$ (Line $XP_3$) is already constructed in the process.
01.01.2022 10:47
FishHeadTail wrote: Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..? Based on the official solution, copying distance without drawing circle is not allowed. You are only allowed two operations: (1) draw a line through two pre-constructed points; (2) draw a circle with centre being a pre-constructed point and radius being the distance between two pre-constructed points. New pre-constructed points can only be form via intersection of such circles and lines. For those curious, the original question has an additional line at the start making reference to the straightedge and compass problem, so it assumes the reader knows the standard rules of the game :p Did a video solution here: https://www.youtube.com/watch?v=n1DOunWAyyk
14.02.2024 00:40
China MO is wack. Call our points $(0,0)$ and $(1,0)$. 1. Draw the x-axis 2. Draw circle centered $(1,0)$ with radius $1$ to mark point $(2,0)$. 3. Draw circle centered $(2,0)$ with radius $1$ to mark point $(3,0)$. 4. Draw circle centered $(3,0)$ with radius $3$ to mark point $(6,0)$. 5. Draw circle centered $(6,0)$ with radius $6$ to mark point $(12,0)$. 6. Draw circle centered $(12,0)$ with radius $12$ to mark point $(24,0)$. 7. Draw circle centered $(12,0)$ with radius $10$ to mark point $(22,0)$. 8. Draw the radical axis of the circles in steps 2. and 3. to mark $(1.5,0)$ 9. Draw the circle centered at $(1.5,0)$ with radius $22.5$ to mark $(-21,0)$ 10. Draw the circle centered at $(24,0)$ with radius $2$ and intersect with the previous circle to get a point that is $\sqrt{2021}$ away from $(-21,0)$
14.02.2024 09:20
Pretty easy problem Construction brings to mind the methods to multiply a segment. To multiply a segment directly by a factor $a$, we must construct $a-1$ circles. Note that we can 'copy the segment' by marking points a unit away from each other. The length $\sqrt{2021}$ immediately gives us the idea of using the Pythagorean theorem. The most obvious way to do this is note $2021 = 45^2 - 2^2$. Copy the unit segment $45$ times and $2$ times, such that we now have marked a distance $A=45$ with endpoints $A_1, A_2$ and $B=2$. Now construct the perpendicular bisector of $A$ and set it to the center of a circle with diameter $45$. At this point we have constructed $2$ lines and $4$ circles. Now draw a circle with radius $2$ at $A_1$ and let its intersection with the larger circle be $X$. Then $A_2X$ will have our desired length of $\sqrt{2021}$ and we are done! $\blacksquare$