# How do you find the center of a circle with a pencil and a book?

Given a circle on a paper, and a pencil and a book. Can you find the center of the circle with the pencil and the book? • Open the book, maybe the answer to your question is written somewhere..! Oct 2 '12 at 17:45
• Math answers should be completely general. Most of those submitted will not work if the circle is much larger than the book. But the book edge can be used to extend the normal to the chord.
– user43409
Oct 2 '12 at 21:56
• offer the book to someone with a measuring device. give them the pen to mark the spot... Oct 2 '12 at 23:25
• can't you just enclose a square around the circle, by using the corner of the book, and then the intersection of the diagonals of the square would be the centre of the circle?
– wim
Oct 4 '12 at 2:52
• @wim This is also a good solution!
– zdd
Oct 4 '12 at 23:28

Let the corner of the book just touch the circle's edge. You can draw two perpendicular lines along book's edge which intersect with the circle at two different points. Connect the two points and it's a diameter.

Repeat this and get another diameter. The intersect of these two diameters is the center of the circle. • very good! any way else?
– zdd
Oct 2 '12 at 12:19
• @zdd You should not be so quick on the trigger with your Approved answer click. This discourages others. Oct 2 '12 at 14:28
• Could some one explain how those two points are the ends of diameter? Oct 2 '12 at 15:23
• See Thales' Theorem. Oct 2 '12 at 15:32
• You assume that book corner are 90 degres. Fair enough. Elegant solution! Oct 3 '12 at 4:10

Not as simple as the accepted answer, but you have another solution:

Use the book to draw two parallel lines that cross the circle at different distances from the center to obatain 4 intersection points in the circle. Draw 4 lines connecting these points. The two points of intersection from these 4 lines define a diameter. Repeat the process with two parallels with different direction to obtain another diameter. Finally the intersection of two diameters returns the circle center. • I added an image to explain. The green line is a diameter because is perpendicular to the parallel lines (in red) and is crossing them at equal distance from both outer segments (in blue).
– Luis
Oct 3 '12 at 2:21
• The picture makes it make a million times more sense. This one gets a +1. Oct 3 '12 at 3:56
• How do you use the book to draw two parallel lines?
– user856
Oct 3 '12 at 3:56
• @RahulNarain, hold the book so it's spine is on the circle, not it's face. But it would work even if the red lines are on opposite sides of the circle. Oct 3 '12 at 3:58
• @RahulNarain The corner of book can be used to draw two parallel lines :) Oct 3 '12 at 8:47

If the circle is a foldable material (like paper), you could fold it in half twice, the intersection of the folds will indicate the center.

• Yes you can fold it, but how can you make sure you are folding along the diameters?
– zdd
Oct 2 '12 at 14:34
• I stand corrected, it would require the circle to be an object by itself, not something printed on a larger body. Oct 2 '12 at 14:58
• +1 you can see through most paper if you hold it up to the light. Oct 3 '12 at 5:11
• How to rescue this: Suppose the circle is printed on stiff, opaque boxboard. Tear a page out of the book (let's assume the pages are reasonably thin and translucent). Using the pencil, trace the circle onto the page. You don't even have to trace the whole circle, just mark off four diametrically opposed arcs long enough for good alignment. Then do the folding trick with this copy. Make a hole in the center of the copy, line up with the original circle and then mark the center through the hole.
– Kaz
Oct 3 '12 at 8:31
• Combine with the Thales solution: Fold one corner of the paper onto the circle perimeter and mark off the two edges with the pencil. Repeat with another corner of the paper. Proceed as above by intersecting the diameters. Question: Is there any situation of a circle on a rectangular paper, where you cannot use the (folded) paper edges themselves as ruler for this construction? Oct 3 '12 at 12:49

As pointed out, most of these answers will not work if the circle is much larger than the book. Instead, find someone with the proper size compass and straightedge, T-square, or other drafting tools, and threaten to hit them with the book if they do not find the center of the circle. See BAROMETER

Meanwhile, what is actually at stake here is Poncelet-Steiner

• +1 for pulling out Poncelet-Steiner to explain, if nothing else, why the book having a corner is critical. Oct 2 '12 at 23:11
• @Will Jagy, thanks for the link, but I think even the book can not cover the circle, it still works, since we can extend the line with the edge of the book. the key point is then the book has the corner as right-angle.
– zdd
Oct 3 '12 at 0:53
• Everyone's a critic. Oct 3 '12 at 2:31
• +1 for the BAROMETER link. And regarding size of book and pencil and circle: the OP provided the pictures of everything, and it's possible to observe that the circle is smaller than the book and the pencil... Oct 5 '12 at 12:56
• No, Poncelet-Steiner is not the solution since it requires a known center. The center in this question is unknown, so Thales' Theorem is more likely.
– spex
Jun 18 '14 at 19:43

Assuming the paper contains nothing but the shape of the circle:

 1. Put eraser end of pencil on book. Or put pencil flush against side of book.
2. Rest circle on tip of point.
3. Circle will rest on the center. • That seems not very practical, but it's for sure something creative! +1 Oct 5 '12 at 8:56
• Hey, some people make a living doing this. youtube.com/watch?v=IRkZN27Hp_k Mar 25 '14 at 3:31

The only book I have is usually under my pillow. It has become skewed over the years - no 90 degrees joy. As a consequence I failed to have success applying the (elegant) description of Patrick Li. Moreover, my book is too small anyway to connect diametrically opposing points on the circle.

Therefore I had to revert to a more tedious approach.

Pick two points on the circle, close enough for my book's reach, connect them. Align one of the book's edges with that line, the bookcorner at one of the marked points, draw a line along the adjacent edge of the book - which is non-perpendicular to the first line. Flip the book and draw "the other" non-perpendicular line through the same point. Repeat at the other point. Connect the two intersections that the four lines make, and extend this line - by shifting one bookedge along - to complete the diameter. If the intersections are too wide apart for my book, retry with the original points closer together. Repeat all of this with two other points on the circle, to get two diameters crossing at the midpoint.

If, by some miracle, my book has straightened out again (but still is tiny compared to the circle), I will quickly find out: the two "non-perpendicular" lines at one of the initial mark-points overlap. Then, I continue the line(s) until I cross the circle at the other side; likewise at the other initial point, and I wind up with a long and narrow rectangle. Repeating that from some other location on the circle, roughly at 90 degrees along the circle from the first setup, I get two rectangles and their intersection is a small parallellogram somewhere in the middle. It's diagonals cross at the center of the circle.

Draw a right triangle inside the circle, making all vertices touch the circle.
Transform it into a rectangle.
Draw a line joining the opposite corners and you will have obtained the center of the circle

Edit: The idea is very simple:
1. Put a corner of the book touching the circle from the inside in any place.
2. Use the edges of the book for drawing the catheti of the right triangle until they intersect the circle.
3. Draw the hypotenuse using those intersections.
4. Put a corner of the book in any of those intersections overlapping one edge with the cathetus.
5. Use the other edge of the book to draw a line until it intersects the circle.
6. Draw a line joining this new intersection with the intersection opposed to the hypotenuse (right angle point).
7. The point in which this line crosses (intersects) the hypotenuse is the center of the circle.

Edit: In fact, any rectangle inside the circle with every corner touching the circle will do the trick

• This would work, but you should specify how you intend to do each step using only the tools available to you. Oct 3 '12 at 20:54
• Thank you Rick. I edited my post following your suggestion Oct 4 '12 at 4:08

# The pencil nor the book are necessary.

Given a circle on a paper

You do the following

1. Fold the paper so only a perfect semicircle is seen through the paper.
2. Unfold the paper
3. Rotate it
4. Repeat Steps 1 and 2

The intersection of the creases will be the center of the circle.

Another way to solve the problem is to pick 3 distinct points on circle say A, B, C. Then connect AB and BC. Halve A-B and B-C to obtain D and E. Draw lines through D and E perpendicular to AB and BC. The intersection of these lines will be the centre of the circle.

• If you can't accurately determine the center of a circle, how can you accurately determine the midpoint of a segment? Oct 2 '12 at 23:02
• @user113215 by taking off a page of the book, cutting it to match the size of the line AB, folding the piece of page that matched the size, putting it over the AB line again to mark what was the middle, and doing that again with the BC segment... Oct 5 '12 at 12:29

1> Place your sheet of paper on top of the circle so that the corner just touches the circle's edge. Hold the paper in place, and use a pencil to make a small mark at the exact point where the two edges that meet at the corner cross the perimeter of the circle.

2> Using a ruler or straightedge, draw a straight line from one mark to the other.

3> Place the paper on the circle again with a different orientation. Make the tick marks again and connect these marks with a second pencil line using the ruler or straightedge. The center of the circle is the point at which this line crosses the first line you drew.

Using Thale's theorem and assuming that the corner of the book is 90°:

• let corner of book touch the circle.
• draw two lines along both edges of book touching the circle at the corner.
• mark intersection of those lines with circle
• draw line between those two intersctions: it's a diameter of the circle
• repeat those 4 steps above for another point on the circle touching the corcer of the book
• intersection of both diameters is the center
• This appears to be the same as Patrick Li's answer. Did you have anything else to add?
– robjohn
Oct 3 '12 at 19:50
• @robjohn: Ups, you are right. I didn't check the other answers carefully enough. At least I mentioned in the answer the central theorem (Thale's) and the required property of the book (90° corner).
– Curd
Oct 4 '12 at 9:10

Hang the circle by its top edge on a wall close to a door or corner, draw a line straight down using the book's edge parallel to the door/corner edge. Rotate the circle f.ex 90 deg., draw a new line and you have the center and radius. This technique can also find the center of more complicated shapes. Gravity is the keyword here.

• This solution uses additional tools, namely the ability to draw a line in a prescribed direction and the ability to find an apex point of the circle, neither of which is really in the (idealized) problem as specified. Oct 2 '12 at 16:58
• @StevenStadnicki No just gravity as a reference. Oct 2 '12 at 19:25
• This answer is nonsense. What it finds is the center of gravity of the object onto which the circle is printed. It assumes that the circle is actually a disc object of uniform density, whose center of gravity corresponds to its geometric centre. The question clearly states that the circle is on paper, not of paper. Even if it's a cutout, paper is likely going to have uneven thickness so that the center of gravity doesn't correspond to the geometric center.
– Kaz
Oct 3 '12 at 8:34
• I don't understand this answer at all Oct 3 '12 at 11:09
• @Kaz :) ok, but you expect to draw very precise lines using the book and the pencil? The book edge will be a very straight line, without imperfections? The tip of the pencil is sharp enought or has some width associated ? Etc... Oct 5 '12 at 12:24

Let D be the center. Let A and B are points on the circle circumference. Let C be another point in the circle circumference.

Then angle ADB is twice angle ACB. I won't provide the proof. Actually it depends on whether C is between A and B or outside it. The theorem is still true but you also need to compute the angle ADB appropriately.

Now, a diameter is a point where the angle of ADB is 180 degree. That means we need to find a point where angle ACB is half of it namely 90 degree.

How?

That book has a 90 degree corner right? Put the corner in the circle circumference. Let's call it C. Now, take the intersection between the edges of the book to the edges of the circle (excluding C). Let's call them A and B. Now, if D is the center of the circle, we know that angle ADB is 180 degree.

We do not know what ADB is. However, 180 degree is simply a straight line. So if we draw a line between A and B, then D must be there.

If D is not on that straight line, then angle ADB won't be 180 degree. That's because the triangle ADB won't be "trivial". That means angle ABD and BAD won't be 0. By sum of angles in triangles, non zero angles of 2 of the angles in a triangle imply that ADB is not 180 degrees. Hence, we can conclude that point D is in a point AB. That means point AB is essentially a "diameter". It's a line that goes to the center of the circle.

Now repeat the process and draw a bunch of diameters. Of those diameters will intersect at some point.

That point is D, which is your center of the circle.

keep the circle on the book at one corner,so as the two sides of the book become tangents @ 90degrees to the circle. the join the two corners of the by a straight line which will pass thro' the center of the circle. now rotate the circle by few degrees closer to 90 degrees and draw another line again as done before.the intersection point of the two drawn lines is the center of the circle. another method. 1) fold the circle to make exactly half(semi)circle.
2) again fold the semi circle to make an exact quarter circle. 3) open and see the two folds meet @ one point, that is the center of the circle. both methods can get the center of the circle.

• The second way was not true, you have no way to fold the circle exactly half, if you can do that, it means you already get the diameter.
– zdd
Oct 3 '12 at 3:10

Use the fact that the book is a straight edge to draw a straight line that touches the circle at one point. Then use the fact that it has a right angle to draw a line perpendicular to that line at that point (where it intersects the circle). This will be a diameter. Repeat this at another point on the circle (not antipodal) to get another diameter. They will intersect at the center.

Follows from the basic fact that any tangent line to a circle is perpendicular to the diameter that goes through the same point (and also from the fact that a standard book is rectangular in shape).

Using just the pencil

1. Mark a point A on the circle
2. Put the beggining of the straight edge of the pencil over the point A
3. Rotate the pencil, keeping the pencil's tip over the point A and the point of your finger on the pencil, accompaining the circle, so that you're measuring the distance from the point A to the point that the pencil touchs the circle
4. Immediatlly before this measure starts to decrease, you've found the circle's diameter.
5. Draw a line connecting point A and this point found on step 5. You've draw the circle's diameter.
6. Choose another point B. Do the same.
7. Where both line intersect, you have the center.

and as bonus:

1. enjoy that you have completed the task and read the book.

A silly answer, im joining to math.stackexchange.com only to answer this, but it may help!

1. open the book
2. put the circle in the middle
3. close the book and see if both parts of the circle are matching (like a mirror)
4. pencil is not needed

valid?

• Cute, but how do you put the circle in the middle (not to mention that you'd have to do this twice, to get two diameters)? Oct 11 '12 at 1:32