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This problem was given to high school students by the Russian prime minister Mishustin.

We have a circle. We are given some point on the circle and its diameter, as shown below. We are given a straight edge and no other instruments. How can you draw a perpendicular line from the given point to the diameter?

enter image description here

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    $\begingroup$ This isn't a puzzle, it's basic geometry. $\endgroup$ Nov 27, 2023 at 3:12
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    $\begingroup$ it's a nice problem $\endgroup$ Nov 27, 2023 at 4:04
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    $\begingroup$ In this case, given that straightedge (and compass) construction is a standard type of maths problem, and the source of this particular problem was given (and can be found e.g. here ), I don't think that changing the description to fix an unintended loophole is wrong. I do feel however that this is more of a maths problem than a puzzle and should therefore be closed. $\endgroup$ Nov 27, 2023 at 10:01
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    $\begingroup$ (Compass and) straightedge construction problems don't have any set procedure by which they can be solved; creative thinking, exploratory approach and an a-ha moment are almost always required, so in my opinion, not only do they count as puzzles, they are often very good ones. $\endgroup$
    – Bass
    Nov 27, 2023 at 17:28
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    $\begingroup$ Thank you @Bass $\endgroup$ Nov 27, 2023 at 22:19

5 Answers 5

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Here's my go (click to embiggen)

enter image description here

Steps:

Connect A to P and pick an arbitrary point Q between them, near-ish to P.

Then, draw lines as shown, constructing the points in alphabetical order, which correspond to colours in the brightening order of black-red-blue-green.

Notice in particular how each new point is at a clearly defined intersection, and that we're only using the straightedge to draw lines through two previously constructed points, just as we should.

Once we're done, point V will be opposite P, and can be used to draw the perpendicular. (Dotted line)


The key points of why this works are

  1. The angles ARB and APB are both right angles, because they are inscribed angles subtended by the circle diameter.

  2. Therefore, Q is the orthocentre of triangle ABS: the right angles mean that AP and BR are both altitudes of triangle ABS, and all altitudes of a triangle meet at the orthocentre.

  3. Therefore, the line through S and Q (blue) must also be an altitude of triangle ABS, and thus, ST is perpendicular to AB.

  4. Using the diameter AB and the newly found perpendicular ST, it is now easy to mirror P to V. (green)

To reconfirm that this works for all points on the circle, we only need to notice that the horizontal position of the perpendicular ST doesn't actually matter; we can use this same ST for any other point P' on the circle, and we only need to construct two new points (U' and V') in order to find the new perpendicular at P'.


EDIT: In the comments, @FlorianF suggests I may have missed a more direct approach by drawing a line through R and the intersection point of AB and ST, and what do you know:

enter image description here

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    $\begingroup$ I'm guessing this was the intended solution. Well done. $\endgroup$
    – msh210
    Nov 27, 2023 at 19:37
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    $\begingroup$ Did you notice how the base of the blue orthogonal lies suspiciously aligned with R and v? $\endgroup$
    – Florian F
    Nov 27, 2023 at 22:21
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    $\begingroup$ @FlorianF Well if that isn't interesting, wasn't a coincidence after all. Good call! $\endgroup$
    – Bass
    Nov 28, 2023 at 1:35
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    $\begingroup$ @FlorianF Not a coincidence, indeed. the incenter (the center of the inscribed circle) of the orthic triangle △DEF is the orthocenter of the original triangle $\endgroup$ Nov 28, 2023 at 8:57
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    $\begingroup$ I didn't think it was a coincidence, actually. But I couldn't prove it. That is why I only hinted at it. Thanks for the link, which helps to prove it. $\endgroup$
    – Florian F
    Nov 28, 2023 at 10:01
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You could fold the paper along the diameter, then mark the mirrored image of the red dot on the lower semicircle, and then connect the two points.

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    $\begingroup$ Now that is thinking outside the box! Brilliant answer! $\endgroup$ Nov 27, 2023 at 11:37
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Note: This is not a valid answer, given the poster's clarification on what the straightedge is capable of. I'm leaving it up because I think it's interesting.


I found a way to solve this, based on

Drawing tangent lines: enter image description here Here's a GeoGebra link where you can play with the construction: https://www.geogebra.org/geometry/sr2ra7jk

To perform this construction,

Use your straightedge to draw the line tangent to the circle through the given point, point A.
Next, use your straightedge to extend the given diameter until it intersects the tangent line. Call the intersection point B.
Next, use your straightedge to draw the line tangent to the circle through B. Let C be the the point at which the tangent line meets the circle.
Finally, use your straightedge to draw the line through A and C. This line is perpendicular to the diameter of the circle, because A and C are symmetrical over the diameter.

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    $\begingroup$ The first two steps will only produce approximate tangent lines, because they're just being chosen by eyeball. If eyeballing were permitted, we could just eyeball the perpendicular directly. $\endgroup$ Nov 27, 2023 at 18:59
  • $\begingroup$ @GregMartin If it's a physical circle and ruler, the exact tangent is found by simply laying the straightedge on the curve. Even if non-physical we're already requiring that we can perfectly align the straightedge to a point, anyway (otherwise you couldn't even extend the diameter), I don't see why you can't do it for a tangent. A bigger issue is that this procedure cannot be implemented for all configurations - when AC is itself a diameter, point B recedes to infinity, making it impossible to draw BC. $\endgroup$ Nov 27, 2023 at 19:20
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    $\begingroup$ "Laying the straightedge on the curve" is eyeballing. $\endgroup$ Nov 27, 2023 at 19:56
  • $\begingroup$ Then drawing a line thru a point is also eyeballing. You typically leave a small gap between the point and the ruler to account for the thickness of your pencil. Drawing a tangent is less precise but not fundamentally different. $\endgroup$
    – Florian F
    Nov 27, 2023 at 21:39
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    $\begingroup$ The entire point of all the compass-and-straightedge construction nonsense is to model Euclid's postulates, so I'm afraid anything more complicated than drawing a line through two points and extending a line will have to be ruled out of bounds. $\endgroup$
    – Bass
    Nov 27, 2023 at 21:40
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Using the ruler, measure the distance to one end of the drawn diameter. Measure the same distance from the same end of the diameter to the other semicircle and draw a point there. Connect the two points with a straight line segment using the ruler.

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    $\begingroup$ In this kind of problems, the operation you do here (essentially marking points equidistant to the distance between two points; then intersecting it with a circle) is supposed to be done using a compass, not a straight edge / ruler. With straight edge / ruler, the only operation supposed to be done is pick two points and draw a straight edge between them. But of course, it's up to Dimitrij to clarify the question :) $\endgroup$
    – justhalf
    Nov 27, 2023 at 5:30
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    $\begingroup$ sorry it's just straight edge, not ruler and no compass $\endgroup$ Nov 27, 2023 at 6:01
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A fairly simple solution is to place the compass on the point, and draw a circular arc (of arbitrary diameter) which intersects the diameter (possibly extended, or "produced", to use the quaint term, outside the circle) at two points.

Then at each of those points draw a couple of circular arcs of equal radius which intersect each other at two points. Then a line through the latter pair of points is the required line.

If a "classical" compass is being used, i.e. one which snaps shut when its pivot point is lifted off the page (unlike a so-called "rusty" compass which stays fixed when lifted and can thereby be used to transfer line segment lengths) then the final pair of arcs can be drawn to pass through the original point on the circle, i.e. using that as a reference length.

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    $\begingroup$ what compass ? only instrument is a straigh edge $\endgroup$ Nov 28, 2023 at 9:22
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    $\begingroup$ Ah yes, I didn't read the question completely. I thought the problem seemed a bit trivial for the elaborate solutions people have given! :-) $\endgroup$ Nov 28, 2023 at 9:28
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    $\begingroup$ I think that, in the world of geometrical constructions, a "rusty" compass is one that has a fixed radius we can never change - so it cannot transport arbitrary lengths, just produce copies of a fixed length. I don't know if there's a standard term for the compass that snaps shut vs. one that doesn't. $\endgroup$ Nov 29, 2023 at 18:13

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