Drawing a unique triangle

Geometry Level 5

Given the orthocenter and circumcenter of a triangle along with a line on which one of the sides of the triangle lie, how many moves are required to draw that triangle using a straight-edge and collapsible compass ?


The answer is 15.

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2 solutions

Sharky Kesa
Mar 26, 2017

I have a 15 move solution.

Let H H and O O be the orthocentre and circumcentre respectively, and the line as l l .

First, we will draw perpendiculars from O O and H H onto l l . The method for drawing perpendiculars from a point P P onto a line p p is as follows:

  1. From an arbitrary point Q Q on p p , draw a circle passing through P P . (1 move)

  2. From another arbitrary point R R (different from Q) on p p , draw a circle passing through P P . (1 move)

  3. These two circles intersect at P P and another point P P' . Draw P P PP' . This is perpendicular to p p . (1 move)

Therefore, it takes 3 moves to construct a perpendicular from an exterior point onto a line, so 6 moves will be used to construct the two perpendiculars.

Now, we will locate the centroid (call it G G ). By the property of the Euler line, we have H G : G O = 2 : 1 HG : GO = 2 : 1 . Thus, we will trisect O H OH partially.

  1. Draw O H OH (and extend it). (1 move)

  2. Draw a circle centred at O O passing through H H . Let this intersect O H OH again at X X . (1 move)

  3. Draw a circle centred at H H passing through X X . Let this intersect O H OH again at Y Y . (1 move)

  4. Draw a circle centred at Y Y passing through O O . Let this intersect the circle centred at O O passing through H H at Z Z . (1 move)

  5. Draw a circle centred at Z Z passing through O O . This intersects O H OH again at G G . (1 move)

Thus, we have the centroid G G located. Let the foot of the perpendicular from O O l l be called M M and the foot of the perpendicular from H H to l l be called N N . Thus, G M GM is the median of the triangle. We will now locate the apex vertex.

  1. Draw G M GM . Let it intersect H N HN at A A . (1 move)

Thus, we have located the apex vertex. We will finally locate the other two vertices and draw the sides.

  1. Draw a circle centred at O O passing through A A . Let it intersect l l at B B and C C . (1 move)

  2. Draw A B AB and A C AC . (2 moves)

Thus, we have drawn the triangle, and it took 6+5+4=15 moves.


Jon Haussman has also got 15 moves:

Construct M M and N N as usual; this takes 6 moves. This construction also gives us line H N HN .

\bullet Draw the circle centred at H H with radius H O HO . (1 move)

\bullet Draw the circle centred at O O with radius H O HO . (1 move)

\bullet Draw the line H O HO . (1 move)

\bullet Let the two circles just drawn intersect at X X and Y Y . Draw X Y XY , and let it intersect H O HO at P P . Then P P is the midpoint of H O HO . (1 move)

\bullet Draw line M P MP , and let it intersect line H N HN at Q Q . (1 move)

\bullet Draw the circle centred at Q Q , and let it intersect line H N HN at H H and A A . Then A A is a vertex of the triangle we seek. (It is easy to see that A H = 2 M O AH = 2MO , which is what we want.) (1 move)

\bullet Draw the circle centered at O O with radius O A OA ; let this circle intersect line l l at B B and C C . (1 move)

\bullet Draw lines A B AB and A C AC . (2 moves)

The total is 15 moves.

A A s e l f self e x p l a i n i n g explaining s o l u t i o n solution w i t h o u t without w o r d s words . S o So D o n e Done i n in 15 15 e l e m e n t a r y elementary c o n s t r u c t i o n s constructions .

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