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
FishHeadTail
22.12.2021 15:50
Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..?
cadaeibf
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).
YaWNeeT
22.12.2021 21:19
Denote by $B(x,r)$ the circle of radius $r$ centered at point $x$. Assume the two given points are $(0,0)$ and $(1,0)$.
(1) Connect the two given points with a line.
(2) Draw $B((1,0),1)$ and find $(2,0)$.
(3) Draw $B((2,0),1)$ and find $(3,0)$.
(4) Connect the two intersections of $B((1,0),1)$ and $B((2,0),1)$ and find $(1.5,0)$.
(5) Draw $B((3,0),3)$ and find $(6,0)$.
(6) Draw $B((6,0),6)$ and find $(12,0)$.
(7) Draw $B((12,0),10.5)$ and find $(22.5,0)$.
(8) Draw $B_1=B((22.5,0),22.5)$ and find $A_1=(45,0)$.
(9) Draw $B_2=B((0,0),2)$ and assume $B_1$ intersects $B_2$ at $A_2$.
(10) Connect $A_1,A_2$. The distance between $A_1A_2$ is $\sqrt{45^2-2^2}=\sqrt{2021}$
CANBANKAN
23.12.2021 04:40
WLOG the two points of distance 1 are horizontal. Call the points $(0,0)$ and $(0,1)$.
By positioning the compass and pencil such that their distance is 1, we can mark $(0,2),\cdots,(0,45)$ with one extra line passing through $(0,0), (0,1)$. (tell me if this is not allowed)
Draw a circle centered at $(0,22)$ with radius 22 and a circle centered at $(0,23)$ with radius 22. Say they intersect at P, Q . Draw PQ and let K be the intersection of PQ and the x-axis. Note $K=(0,\frac{45}{2})$.
Now draw a circle centered at $K$ passing through $(0,0)$.
Draw a circle centered at $(0,45)$ with radius 2. The intersection of this point with the circle centered at $(0,\frac{45}{2})$ has distance $\sqrt{2021}$ from the origin.
FishHeadTail
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?
CANBANKAN
23.12.2021 10:20
No, because it doesn't create a new circle or line.
cadaeibf
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.
CANBANKAN
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.
FishHeadTail
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 $10$.
Let $P_0, P_1$ be the two given points. In the following construction, $R-S$ denotes drawing a line through $R, S$ (using the ruler once) and $C(R, RS)$ means to draw a circle with center $R$ and radius $RS$ (using the compass).
1) $P_0-P_1$. Call this line $l$.
2) $C(P_1, P_1P_0)$ and call this $C_1$.
....Let $C_1$ intersects $l$ at $P_0, P_2$, so $P_1P_2=1$.
3) $C(P_2, P_2P_0)$ and call this $C_2$.
.....Let $C_2$ intersects $l$ at $P_0, P_3$, so $P_2P_3=2$.
4) $C(P_3, P_3P_1)$ and call this $C_3$.
.....Let $C_3$ intersects $l$ at $P_1, P_4$, so $P_3P_4=3$.
5) $C(P_4, P_4P_1)$ and call this $C_4$.
.....Let $C_4$ intersects $l$ at $P_1, P_5$, so $P_4P_5=6$.
6) $C(P_5, P_5P_2)$ and call this $C_5$.
.....Let $C_5$ intersects $l$ at $P_2, P_6$, so $P_5P_6=11$.
7) $C(P_6, P_6P_1)$ and call this $C_6$.
.....Let $C_6$ intersects $l$ at $P_1, P_7$, so $P_6P_7=23$, meaning that $P_2P_7=45$.
* 8) $C(P_1, P_1P_4)$ and call this $C_7$.
.....Let $C_7$ intersects $C_4$ at $A, B$.
* 9) $A-B$ and call this $l’$.
.....This is a line perpendicular to $l$ at $P_3$.
10) $C(P_2, P_2P_7)$ and name this $C_8$.
.....Let $C_8$ meet $l’$ at $X$.
Now $XP_3=\sqrt{2021}$ as $2021 + 2^2=45^2$.
CANBANKAN
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!
YaWNeeT
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 $10$.
Let $P_0, P_1$ be the two given points. In the following construction, $R-S$ denotes drawing a line through $R, S$ (using the ruler once) and $C(R, RS)$ means to draw a circle with center $R$ and radius $RS$ (using the compass).
1) $P_0-P_1$. Call this line $l$.
2) $C(P_1, P_1P_0)$ and call this $C_1$.
....Let $C_1$ intersects $l$ at $P_0, P_2$, so $P_1P_2=1$.
3) $C(P_2, P_2P_0)$ and call this $C_2$.
.....Let $C_2$ intersects $l$ at $P_0, P_3$, so $P_2P_3=2$.
4) $C(P_3, P_3P_1)$ and call this $C_3$.
.....Let $C_3$ intersects $l$ at $P_1, P_4$, so $P_3P_4=3$.
5) $C(P_4, P_4P_1)$ and call this $C_4$.
.....Let $C_4$ intersects $l$ at $P_1, P_5$, so $P_4P_5=6$.
6) $C(P_5, P_5P_2)$ and call this $C_5$.
.....Let $C_5$ intersects $l$ at $P_2, P_6$, so $P_5P_6=11$.
7) $C(P_6, P_6P_1)$ and call this $C_6$.
.....Let $C_6$ intersects $l$ at $P_1, P_7$, so $P_6P_7=23$, meaning that $P_2P_7=45$.
* 8) $C(P_1, P_1P_4)$ and call this $C_7$.
.....Let $C_7$ intersects $C_4$ at $A, B$.
* 9) $A-B$ and call this $l’$.
.....This is a line perpendicular to $l$ at $P_3$.
10) $C(P_2, P_2P_7)$ and name this $C_8$.
.....Let $C_8$ meet $l’$ at $X$.
Now $XP_3=\sqrt{2021}$ as $2021 + 2^2=45^2$.
The problem asks for a line and two points on it, so I think you need one more line $XP_3$.
FishHeadTail
25.12.2021 11:28
$l’$ (Line $XP_3$) is already constructed in the process.
renrenthehamster
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
AdventuringHobbit
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 $(-21,0)$
IbrahimNadeem
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! $\blacksquare$