## More maths – A little geometry

Posted by Karys on June 17, 2010 – 2:08 pm

Let ABC be a *direct* triangle (you go from A to B to C counterclockwise) in a plane.

D,E,F,G,H five points in the same plane, such that :

-ABDE is an *indirect* square.

-BCFG is an *indirect* square.

-BDHG is a *direct* parallelogram.

Show that **AH=CE**

and that **(AH) and (CE) are orthogonal/perpendicular** to each other.

*(HA, no more “the answer’s [answer]” and finished… here you have to give proof…)*

June 18th, 2010 at 8:08 am

Just to show some ignorance here – what do you mean by an indirect square?

I can draw up a diagram that I can force to meet your criteria, but whether they are indirect squares/parallelograms or not, I don’t know.

Karl

June 18th, 2010 at 8:10 am

Hold on.. i’m calling my brother, he should be able to figure it out……

…. it’s ringing…

June 18th, 2010 at 8:13 am

What I mean is like the triangle.

The square ABCD is indirect if you go from A to B to C to D clockwise.

In fact, a more appropriate definition is with directed angles :

The square ABCD is indirect if (AB;AD)=-pi/2 [2pi].

June 18th, 2010 at 8:13 am

A——B

——–

——–

——–

D——C

ABCD is an indirect square.

June 18th, 2010 at 8:14 am

Rectangle…

June 18th, 2010 at 10:40 am

And a direct parallelogram is?

June 18th, 2010 at 11:16 am

Actually there is only one parallelogram possible in our case when square and triangle are set…

Anyway, for any convex polygon, you just have to replace “square” by “any convex polygon”, and the according points.

Remember that ABCD is the same as BCDA, but not ABDC. You trace the polygon following the order of the points.

June 18th, 2010 at 1:45 pm

Based on direct=counterclockwise and indirect=clockwise I am able to draw the layout per your instructions.

Proving this is another issue.

I drew point A at the top of my triangle with B being directly below it and C directly to the right of B. AB = BC at perfect right angle.

Points E & D lie directly to the left of points A & B respectively so that AB=BD=DE=EA or a square.

Similarly points G & F lie directly under points B & C respectively so that BC=CF=FG=GB or a square.

This causes my parallelogram to also be a square since H must be placed direclty left of G and directly below D which happen to be equidistant from B and in a perfect 90 degree angle.

I agree this is a very special case but nevertheless it is a possible layout and easiest to prove.

In my simple form I have a square with a missing upper right corner. Each segment of the quare has a mid point A, C, D, or G therefore The distance from EC is the same as DF and if you rotate the Square CE overlays HA perfectly.

Not exactly proof since it is difficult to use this method when point C does not lie exactly 90 degrees from AB with AB=BC.

As C moves away from B, square BCFG increases in size, the angle of AEC decreases at the same rate as the angle DHA.

As C moves away from 90 degrees BCFG moves accordingly which casues BDHG to become a recognizable parallelogram. This is where the math gets ugly and beyond my explanation abilities.

Building this with tinkertoys and 2 pieces of string would be a piece of cake. 2 simple squares which pivot on point B and a floating point H which completes the parallelogram. Building it so that point c is flexible requires flexible joints that move in unison.

June 18th, 2010 at 4:08 pm

I confirm that what you drew, even though was a very particular case, is correct.

June 19th, 2010 at 8:43 am

I’m with John 24 here, I can draw it, I can only draw it with a right angle triangle, but the parallelogram will therefore always be a square/oblong – so why not say square/oblong.

Using an RH triangle, the answer is in Pythagoras theorem. Start with a basic 3/4/5 triangle, draw the triangle with a square below the x axis and to the left of the y axis. Join the 2 squares to create an oblong (parallelogram).

The lines AH and CE creat perfect RH Triangles… do the pythagoras math and they are the same length.

June 19th, 2010 at 11:14 am

What do you mean by “I can only draw it with…” ?

June 19th, 2010 at 3:46 pm

Karys, I mean to say that my solution is with a right angled triangle making points ABC. This means that the parallelogram BDHG will be an oblong, and this may not really be the intended solution, unless the use of the words square (defined as a parallelogram with all sides equal in length and all angles fixed at 90°) and then parallelogram which is in fact a rectangle (defined as a parallelogram but with all angles fixed at 90°) was to throw us off track?

I would submit a picture, but I can’t seem to post them here…

June 19th, 2010 at 4:04 pm

Actually, I’m expecting an answer for any triangle ABC — The squares and parallelogram depend on this triangle.

I found this interesting problem in my math book, in the chapter on similarities. That was a hint to solve this problem; maybe there are other ways. Anyway I wanted to see if people on ToM handle plane transformations as well as arithmetic and pure logic.

June 24th, 2010 at 2:12 am

Because it looks like noone’s gonna answer that one, I’ll give the answer tomorrow…

June 24th, 2010 at 3:55 am

The problem with answering the question is that describing the reasoning without a picture is too difficult. It is also clear from the responses that some people haven’t even been able to construct a reasonably general picture of the situation.

Having made the construction, it is easy to see that by (mentally) rotating the whole thing 90 degrees clockwise about the centre of the square ABDE, that triangle ABC will lie exactly over where triangle BDH was and so BDH is congruent to triangle ABC and then it is easy to also see that triangle ECA is congruent to triangle AHB. In fact you can see that ECA is congruent and rotated (by 90 degrees) to AHB directly. So the length of AH = length of CE and they are perpendicular (courtesy of the 90 degree rotation).

The above won’t make sense if the reader can’t draw the original construction.

June 24th, 2010 at 4:07 am

PS introducing non-standard terms such as “direct” (which you did at least define) and “indirect” (which you didn’t define) rather than sticking with “anti-clockwise” and “clockwise”, alienated some people from the problem.

June 24th, 2010 at 7:06 am

Thanks Chris…. I didn’t even look at rotating my diagram… too hung up on angles I guess. I have traced and overlaid my RH triangle based diagram, and lo and behold….

June 24th, 2010 at 8:46 am

In my sketch I made the second square roughly twice as large as the first, and with an approximately 30 degree anti-clockwise rotation with respect to the first square, hence my ABC was obtuse, but I could have made it be acute without changing any of the reasoning.

I hope the following displays ok:

C

A B

E D

F

G

H

June 24th, 2010 at 8:48 am

BCFG is meant to be a clockwise square.

June 24th, 2010 at 10:37 am

Ok, I should have not introduced the “direct” / “indirect”,

It’s just that in France those were the words used, and it didn’t come in mind that “clockwise” and “anti-clokwise” would be more appropriate.

However, Chris answered correctly at last

June 24th, 2010 at 11:57 am

Hi Karys. It’s difficult to advise you, it may have been best to simply state all the shapes in “direct” notation (having mentioned anti-clockwise). i.e. AEDB, ABC, BGFC.

Anti-clockwise is also the standard orientation when doing contour integrals and rotation vectors. I assume this to be an international convention.

June 24th, 2010 at 11:17 pm

Actually, yes it seems to be a convention, which explains why I began with “direct”.