Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Triangle ∆ABC is arbitrary. Side AC is a diameter to a semicircle, and M marks the point in the middle of this semicircle arc. Side BC is a diameter to a semicircle, and N marks the point in the middle of this semicircle arc. O is a point on side AB, and AO = OB.

Conjecture: OM = ON and $\measuredangle$MON = 90°

share|cite|improve this question
One should use the exterior semicircle both times, i suppose – Hagen von Eitzen Jan 1 '13 at 17:28
Just to make sure I understand, point $M$ is on the bisector of $AC$, outside the triangle and a distance $\frac{1}{2} |AC|$ away from $AC$? – copper.hat Jan 1 '13 at 17:31
@Hagen: it seems either the two exterior semicircles or the two interior semicircles, but not a mixture. – Henry Jan 1 '13 at 17:51
Yes, my apologies. Indeed, these are exterior semicircles. – rtukkine Jan 1 '13 at 18:34

With the vertices represented by complex numbers $a,b,c$, we find $m=a+\alpha(c-a)$, $n=c+\alpha(b-c)$, $o=\frac{a+b}2$ with $\alpha=\frac{1+i}2$. Now oberve that $$m-o = -\frac i2a-\frac12 b+\frac{1+i}2c$$ and $$n-o = -\frac 12a+\frac i2 b+\frac{1-i}2c$$ and hence indeed $ m-o = i(n-o)$.

share|cite|improve this answer
Why is alpha (1+i) / 2 ? – rtukkine Jan 1 '13 at 20:14
Also, is there a way to prove this without complex numbers - e.g. using symmetry? – rtukkine Jan 1 '13 at 22:16
@rtukkine: To reach M you can do the following: start at A ($a$), go towards $C$ (i.e., in direction $c-a$) for half the distance (reaching the point $a + \frac12(c-a)$), then take a left turn and go the same distance (reaching $a + \frac12(c-a) + i\frac12(c-a)$). This is the point $a + \frac{1+i}{2}(c-a)$, and in this the above answer has put $\alpha = \frac{1+i}{2}$. The $i$ comes from the left-turn (multiplication by $i$ is counterclockwise rotation by 90°). – ShreevatsaR Jan 2 '13 at 14:36

Let's suppose two exterior semicircles in our approach. See the figure bellow. TriangleSemicircles

Note that CJOK is a paralelogram, that $\angle CJO$ is external of $\triangle JAO$ and that $\angle CKO$ is external of $\triangle KOB$. Note also that $\triangle MJO$ and $\triangle OKN$ are congruent, therefore $$MO = ON.$$ Using $\triangle MJO$ we get: $$\alpha + \beta + 90^{\circ} + \theta + \gamma = 180^{\circ} \Rightarrow$$ $$\alpha + \beta + \theta + \gamma = 90^{\circ}$$ But as $\angle AOB$ is a straight angle and $\alpha + \beta + \theta + \gamma = 90^{\circ}$ we can conclude that $\angle MON$ is a right angle.

You can use a similar approach when semicircles are not exterior.

share|cite|improve this answer
How do you know that CJOK is a paralelogram? – rtukkine Jan 3 '13 at 20:48
@rtukkine. The line segment connecting midpoints of two sides of a triangle is parallel to the third side and its length is half of the length of the third side. – RicardoCruz Jan 3 '13 at 21:24

By "mathematically" do you mean by coordinates as opposed to geometry? By coordinates, you can let $A=(-1,0), B=(x,y), C=(1,0)$ by rotation, translation, and scaling. Then $M=(0,-1), N= (\frac{x+y+1}2,\frac{x+y-1}2)O=(\frac x2,\frac y2)$. Now you can check.

share|cite|improve this answer
Not sure what you mean. – rtukkine Jan 1 '13 at 19:27
@rtukkine: I have shown the coordinates of points $O,M,N$ in the plane. You can now check whether the distances match and whether the angle is a right angle by checking whether the slopes of the sides multiply to $-1$ – Ross Millikan Jan 2 '13 at 2:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.