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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°

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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
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3 Answers 3

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)$.

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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
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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.

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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
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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.

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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
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