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**MrFreeze****Member**- Registered: 2010-11-15
- Posts: 2

I have a series of questions that I have been trying to solve and I need help with the solutions, they are driving me crazy! Each question needs to be proven also, not just constructed. Here they are:

1. Describe all isometries of the Manhattan plane.

2. Two points A and B lie on one side of line ℓ. Two points M and N are

chosen on ℓ such that AM +BM is minimal and AN = BN. Show that points

A, B, M and N lie on one circle.

3. Given two parallel lines ℓ and m and a point P, use only ruler to construct

the line through P parallel to ℓ and m.

4. Given two concentric circles construct their center using only ruler.

5. Show that any construction with only ruler can be done with a short ruler;

i.e. an instrument which makes possible to draw a line only through sufficiently

close pair of points.

6. Assume you have an instrument which makes possible to draw a circle or

line through any given three points. Show that it is impossible to construct the

center of given circle using only this instrument.

7. Give a ruler-and-compass construction of a circle or a line which perpendicular

to each of three given circles. (You may assume any two of three given

circles do not intersect.)

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,072

Hi MrFreeze,

I've only just started to think about these so no answers yet.

Some clarifications needed.

(1) What do you mean by Manhattan plane? The only hits I get from Google involve an aircraft crash so I cannot connect that to isometries.

(2) Using a ruler. I could use it in three ways: (a) to make a straight line (b) to make two lines by using both edges together; these would be parallel (c) to measure lengths.

With (2c) I can find the centre of a single circle so don't need two concentric circles. Do you want this solution or must I persist with straight edges only?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**MrFreeze****Member**- Registered: 2010-11-15
- Posts: 2

sorry, can't use lengths. only straight edges

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,072

Ok Here's my answer to number 4.

It requires a ruler with width lower than the diameter of the smaller circle and smaller than the radius of the larger circle.

Lay the ruler across the circle so that it cuts both circles twice.

Draw parallel lines by using both edges of the ruler.

Label the points where the lines cut the large circle A, B, C, and D and the small circle points E, F, G and H. (My first and second diagrams have swopped over so see second diagram).

Join AC and BD and label the crossover J.

Join EG and FH and label the crossover I.

ABD = BDC ( parallels cut by transversal => alternate angles)

BAC = BDC (angles subtended by same chord)

So AJB is isosceles.

Similarly EIF is isosceles.

IJ is perpendicular to AB (and EF) because (oh drat I forgot to label the point where IJ cuts AB. Let's call it X.)

AJ = JB (radii) BAJ = ABJ (proved above) XJ is common to AXJ and BXJ.

So AXJ and BXJ are congruent so AXJ = BXJ = 90.

So IJ goes through the centre of the circle (perpendicular bisector of chord goes through centre)

Now refer to simplified first diagram.

Let this diameter cut outer circle at K and L

Turn ruler sideways and lay one edge along the diameter.

Draw the parallel line MN cutting the circle at M and N by using the other edge.

and another similar line OP as shown.

Join ML and KN to cut at R.

Join KP and OL to cut at S.

A similar proof to part one shows that the centre lies on SR.

Where SR crosses KL is the required centre. See point T.

Bob

ps I still don't know what a 'Manhattan Plane' is.

*Last edited by bob bundy (2010-11-18 01:47:49)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,072

Here's my solution to number 4.

Draw a line and choose points A and B.

For AN = BN construct the perpendicular bisector of AB and mark N where it crosses the line.

For AM + MB to be minimal, reflect B in the line and call this point B'.

Join AB' and mark M where this line crosses the given line.

The distrance is minimal because if you move M the segments AM and MB' have a bend at M so the distance is longer. As MB' = MB the distance AM + MB is minimised by this construction.

As N lies on the perpendicular bisector of AB, any circle with centre chosen on this Perp. Bis. will go through A and B.

AMB will have a circumcentre, constructed by finding where the perpedicular bisectors of AM and MB meet.

Call this point C. Already, we know AC = BC = MC.

But C must also lie on the perp. bis. that gave N

There are infinitely many circles that can be centred on this line that go through A and B, including one that also goes through N.

It remains to show that the circle that goes through A, B and N is also the circle that goes through A. B and M.

They are, but the proof eludes me at present.

Back later to finish this off !!!

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,072

OK got it. But it's a bit of a chase of angles around the diagram so rather than add it to the above post, I thought it would be easier to show as a separate post. My geometry software insists on showing A, B N and M on the same circle (because they are!) but that makes it hard to avoid making assumptions that haven't been justified. So I started with a new diagram with A, B and M as before but with N deliberately shown not on the circle. My aim is to show it is. (see diagram below)

Let x and y be the angles shown on the diagram.

Then the following angles can be found by angle sum of a triangle, reflection properties, isosceles triangles, straight line etc:

MBC = y

BMC = y

MCB = 180 - 2y

CMA = 180 - 2x - y

MAC = 180 - 2x - y

ACM = 4x + 2y - 180

ACB = 4x

Now we do know that N lies on the perpendicular bisector of AB so

NCB = 2x

NCM = 2x + 2y - 180

CNM = 180 + 180 -2x -2y -x - 180 + 2x + y

= 180 - x - y

NMC = 180 - 2x - y + x

= 180 - x - y

So CNM = NMC

which fixes N on the circle as NCM is isosceles.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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