Just making a point
Prove that the three perpendicular bisectors of a triangle always meet at a point.
Here's a link so you can see what I'm blathering on about:
http://www.analyzemath.com/Geometry/Circumcircle/Circumcircle.html
On that page, click the "click here to start button". Ignore the circle that you see. Drag the triangle around and see what happens.
Classical geometry on allows the use of a straightedge and compass. Rulers and protractors are not allowed. Trigonometry is not allowed.
However for this problem any reasonable proofs are OK with me.
Here's a link so you can see what I'm blathering on about:
http://www.analyzemath.com/Geometry/Circumcircle/Circumcircle.html
On that page, click the "click here to start button". Ignore the circle that you see. Drag the triangle around and see what happens.
Classical geometry on allows the use of a straightedge and compass. Rulers and protractors are not allowed. Trigonometry is not allowed.
However for this problem any reasonable proofs are OK with me.
Labels: mathschallenge, SharedPuzzle
posted by Chris at
7:02 PM
24 Comments:
i dont get u , u want us to roove that the perpendicular bisectors of a triangle meet at the same point? Isnt that obvious? its called the circumcentre.........
Hi Anonymous. It is not obvious. You are simply remembering a related fact which is of similar complexity. Can you prove the related fact?
I suspect that a proof of one may only need a small tweak to get the other.
At this moment, I don't know how to prove either of the facts.
ok, here is my algebric answer:
Suppose the following vertices of the triangle: origin O, complex point A and complex point B.
Each point of the line that bissectes OA (and is normal) can be written of the form R=A/2+k.i.A (that is, passes by A/2 - midpoint - and is directed normal to OA)
The same for the line that bissects OB: R'=B/2+k'.i.B
For AB, the result is R''=(A+B)/2+k''.i.(A-B).
If it is true that those trhee lines pass are concurrent at point P, then there exists a K, a K' and a K'' real numbers such as R=R'=R''. Let's proove it.
R=R'=R'' can be written as a system of 2 complex equations:
A/2+k.i.A=B/2+k'.i.B
A/2+k.i.A=(A+B)/2+k''.i.(B-A)
Whe have a system of two equations and three variables, but remember that K, K' and K'' are real, and if we separete the real and imaginary part of the equations we have 4 equations and 3 variables (suppose here that A=xa+i.ya, B=xb+i.yb):
xa/2-k.ya=xb/2-k'.yb
ya/2+k.xa=yb/2+k'.xb
xa/2-k.ya=(xa+xb)/2-k''.(yb-ya)
ya+k.xa=(ya+yb)/2+k''(xb-xa)
If there is one solution to the above system, then we prove that all the three lines intersect at the same point. To prove this, we must prove that:
- The caracteristic determinant of the system is 0 (so that one of the equatians can be written using the other 3)
- Removing one equation the system is possible.
Using a software for symbolic calculus, both of these are easy to prove (I used Derive 6.0 for windows). The same result cam be achieved "by hand", but involves a little bit of work.
Hi, Chris.
It can be done as any geometrical problem with the help of coordinate method. Let r_1, r_2 and r_3 be vertexes of triangle and wright perpendicular bisectors as follows
u=0.5(r_1+r_2)+n_u*t_u,
v=0.5(r_1+r_3)+n_v*t_v,
w=0.5(r_2+r_3)+n_w*t_w,
where
(n_u, r_2-r_1)=0, (scalar product)
(n_v, r_3-r_1)=0,
(n_w, r_3-r_2)=0.
Intersection point of first two
u(t_u)=v(t_v), hence
0.5*r_2+n_u*t_u=0.5*r_2+n_v*t_v
Then perform scalar multiplication of both sides by r_2-r_1. You'll get as a result
t_v=((r_2, r_2-r_1)-(r_3,r_2-r_1))/(2*(n_v,r_2-r_1)).
Then find intersection point
v(t'_v)=w(t'_w)
using now scalar product with r_2-r_3. You will get t'_v formula with apparently equal nominator and also equal denominator, the last is shown with the help
of (n_v, r_3-r_1)=0. Hence intersection points with parameters t_v and t'_v are the same.
LOL. Miguel, you read my mind. I had a pretty shrewd idea that you were going to try it that way. It was in the back of my mind when I posted the problem. I started to develop an answer on those lines, but realised that the simultaneous equation stuff was going to make the proof rather unwieldy to post. Thank you so much for rising to it.
I'd like to see a more conventional proof though.
OK, I have another one solution. Now geometrical. But I need a figure, so repeat my instructions. Draw a triangle ABC. Built perpindicular bisectors on AC and BC sides, correspondingly OD and OE(they intersect in a some point O). The following right triangles are equal
(AOD, ODC), (OCE, OEB) because they got equal legs. Thus AO=OB. Thus if you build perpindicular in AOB from O it will divide AB by two equal parts.
Hi quantense, your first answer wasn't up when I started to reply to Miguel. I, in fact only glanced over Miguel's answer because I knew that at worst he would have only made a typo type of error and I pretty much knew what he was going to go before he'd posted it. I had to go for a coffe before I could respond to you, that's lucky, because you had already posted you second proof when I came back.
Although your first answer used vectors and scalar products, it looks like a good answer. In fact I was considering relenting and permitting solutions using vector methods.
Although my brain had to strain a bit with your second answer, I think you have a very nice pure geometric proof, and I suspect that it is about as good a proof as one could wish to hope for. Thank you very much.
... I will look at all the solutions more thoroughly. I wouldn't like anyone to think that their work wasn't being apppreciated.
This post has been removed by the author.
ok, the simplest I can manage:
let the vertices be V1=(0,0); V2=(0,C); V3=(A,B)
the equation of the line that bissects V1V2 is easy:
y1=-A/B.x+B/2+A^2/(2B)
the equation that bissects V2V3 is also easy:
y2=-(C-A)/B.x+B/2+(C-A)^2/(2B)
These two lines intersect, when y1=y2, at a point P with x=C/2 (solve the equality). So, the line that bissects V1V2 (which is on the xx axis) passes by this point P, QED
:)
Me wittering on again:) The title of the problem had a double-meaning. I was trying to give an (admittedly poor) definition of geometry, as I'd noticed that some posters aren't clear about it.
Rulers, protractors and trignometry are not allowed in pure geometry. Only compasses and straightedge (and pencil) are allowed.
I can try other proofs if I have more time, but it is dificult to get simpler then this last one. The trick is the placement of the vertex, so that all the algebra is very simple.
Got a simpler one... it's incredible how I didn't remember this before; my brain must be very complicated... :)
Givem three points A, B and C non colinear, it is always possible to draw a circumference that passes by the three points (easy). If P is the center of that circunference, then P is equally distant from A, B and C (by definition), and so the line that connects the midpoint of AB to P is normal to AB, and the same to AC and BC. Thus, P is the point we are looking for.
Hi Miguel, your last proof has the drawback that finding the equation for a line perpendicular to a given line passing through a given point is required. I'd have to look that up (or work it out). But I wouldn't call it a serious drawback, unless those perpendicular line equations required trigonometry - I imagine that trig could be avoided though.
I guess you have a typo though, the perpendicular bisector of V1 V2 is y1 = C/2. But I'm pretty sure the only fault(s) are typos.
Head stating to hurt, I haven't done co-ordinate geometry for decades, at least not of this type.
Chris, the slope of two perpendicular lines are related as m=-1/m'. That's very easy to demostrate using vector dot product.
...and no, no typos. The point P has x coordinate C/2, and that's why the (vertical) line that bissects V1V2 (which is horizintal) passes at it.
Miguel, that was excellent. I can now say that the result is obvious. LOL. (see below).
I feel that I should partially apologise to the first anonymous poster for my response. The reason that I couldn't accept your answer is that you didn't give enough of an explanation. At the time, I couldn't see what you were driving at - I'm not telepathic. Although I'm guilty of using the word "obvious" myself sometimes, I often find that when other people use it, their obvious isn't my obvious (or something like that).
Greetz all.
Hi Miguel. I was pretty sure (99.9999999%) that that's how the slopes related, but I seem to recall that it used trig. i.e. tan(90+A) = -1/tan(A). Soz if I'm guilty of seeing typos if they're not there. But my head really is slow/lazy today.
Miguel, not sure if you saw it, but until the "bug on a rubber band" problem came along, I thought that the "Little Two" was one of the nicest problems I'd seen. It was posted on 16/7/09. I had posted a good solution. Unfortunately the layout went wonky as my lines were too wide. I'd do a more concise solution now I've got warmed up over the last few months. Greetz.
Some of these post are getting mixed up because we're posting in parallel. SO a reference to "last post" actually meant "before the last post".
Chris, I solved the "little two" problem, but your answer is much more elegant. I posted there.
Hi Miguel, your "Little Two" post hasn't appeared. I've noticed the blogger software is a bit buggy.
Miguel, I've found that when a post doesn't appear, posting something else to the same blog makes them both appear.
I'm a nut. Co-ordinate geometry is accepted as part of modern (cf classical) geometry. I absolutely assume that the -1/m gradient can be argued by geometric construction.
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