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**tony123****Member**- Registered: 2007-08-03
- Posts: 205

find AB

FROM Mr Doloros

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

Hi tony123;

*Last edited by phrontister (2020-04-08 15:55:17)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,930

hi

Great problem. Thanks, Tony.

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

Hi Bob;

*Last edited by phrontister (2020-04-07 12:56:28)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,930

hi phrontister

Hhmm. See pic:

The diagram doesn't say that DM is 4. Nor does it say that MKA is a straight line. Maybe it is (looks like it) but I'm still thinking you've made assumptions that aren't there.

Unless you care to prove these. (Trouble is you're right … but I'm being picky. There's lots of fallacies in maths made because something 'looks right'.)

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

Hi Bob;

The only dimensions given by the OP are those of 3:4:5 triangles, and therefore I strongly suspected that the solution would involve such triangles. (barring red herrings, that is!)

So I drew the LABC rectangle, which led to the image below (Geogebra) and my post #2 solution.

That's probably not a proper proof as such, but I've failed to find one of those things, though I've spent some time trying.

And I can't find any other solution...

*Last edited by phrontister (2020-04-08 16:16:20)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,930

hi phrontister

I made my diagram using Geometer's Sketchpad. I've had that program since before Geogebra was introduced so I've stuck with it. I have used Geogebra and even taught classes using it, but I'm more familiar with Sketchpad so I usually use it for problems. I made my diagram like this:

Draw AB of indeterminant length. Construct AL so that BAL = 90 and AL = 4. Centre of a circle at L, radius 3. Choose any point K on the circle and construct a tangent at K. Mark M so that KM = 4. Construct line at M, at right angles to KM.

Not surprisingly, this line did not go through B.

So I picked the point K and moved it around the circle, thereby moving that last line until it did go through B. That gave me the diagram and when I measured AB I got a value of 8.01. Is that close enough to 8 to declare that's the answer?

Well no. Firstly all vector geometry programs have a limit on accuracy. I've noticed Geogebra can work more accurately but even so, a diagram doesn't constitute a genuine proof. There are many fallacies based around poor sketches. What Euclid teaches us is that we must try to put together a proof. So that's what I tried to do.

That LM is parallel to AB and therefore DM = 4, comes out as a property of the original information. You don't have to assume it. Here's my solution:

See earlier diagram.

Construct EKF so that it is parallel to AB, with MFD parallel to AL, and GK parallel to AL with G on AB. That makes three right angled triangles: LEK, KFM and MDB

I have marked angle ELK with letter x. So LKE = 90 - x, FKM = x, FMK = 90 - x, and DMB = x.

So the triangles are similar.

DM = 5cos(x) so DF = DM - FM = 5cos(x) - 4sin(x) = AE

EL = AL - AE = 4 - 5cos(x) + 4sin(x)

But EL = 3cos(x), therefore

3cos(x) = 4 - 5cos(x) + 4sin(x) => 8cos(x) - 4sin(x) = 4

AB = AG + GD + DB = EK + KF + DB = 3sin(x) + 4cos(x) + 5sin(x) = 8sin(x) + 4cos(x) = 4(2sin(x) + cos(x) )

If cos(alpha + x) = 1/root(5) then sin(alpha+x) = 2/root(5)

As trig theory requires similar triangles I have tried to eliminate the trig from this proof; without success so far.

The properties like M on LC, DM = 4 and AKM a straight line, then follow.

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

Hi Bob;

Wow...that's quite some proof!

However, the trig's too advanced for me to understand it...but I'm happy to wait for a simplified solve if you can find one.

I'm glad, though, that your proof confirms that my solution is correct.

bob bundy wrote:

The properties like M on LC, DM = 4 and AKM a straight line, then follow.

Btw, I don't think that AKM is a straight line (see my Geogebra image below).

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**Bob****Administrator**- Registered: 2010-06-20
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Not a straight line. Agreed. In my diagram it looked like a near miss so I assumed it would be right. Refer to signature third line

Bob

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Bob Creamer****Member**- Registered: 2020-04-11
- Posts: 1

Using Bob Bundy's diagram and measuring the distance below AB by tracing through the points ALKMB

we get AL -LE +FM -MD =0

or 4 -3 cos x + 4 sin x -5 cos x = 0

or sin x - 2 cos x = - 1

between 0° and 90° both sin x and -2 cos x are increasing so only one single solution.

And by enlightened guesswork (and phrontister's hint of 345 triangles) we find that

sin x = 3/5 => cos x = 4/5 and this solves sin x -2 cos x =35/ - 2*4/5 = -5/5 = -1

This is a lot simpler than Bob Bundy's method of converting sin s - 2 cos x into K sin (x +α)

and reduces it from what I guess is Further Maths A-level to GCSE

So Yes it is using 345 triangles

And by a similar route to the above

AB = EK +K F + DB

= 3 sin x + 4 cos x + 5 sin x

= 8 sin x + 4 cos x

= 8 x 3/5 + 4 * 4/5

= (24 + 16)/5

AB = 8

*Last edited by Bob Creamer (2020-04-11 12:36:12)*

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**Bob****Administrator**- Registered: 2010-06-20
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hi Bob,

Welcome to the forum.

I got to 8cos(x) - 4sin(x) = 4 , didn't like the look of that so started again with AB = …

When I got to a similar expression I jumped into compound angle formulas like a bloodhound hot on the track of a fugitive. Never occurred to me to just try and find x.

So thanks for your contribution. Some might say "Is enlightened guesswork allowed?" but I'm perfectly happy with it as you have justified the unique solution. It's how we do most integration and no one objects then.

My 'logic' for an even simpler solution is that all trig is underpinned by similar triangles, so a solution that bypasses the trig may be possible. I've been busy with a couple of other problems in the Help Me section over the last few days so it may be a while before I return to this one. When I first joined there was an integration problem that had everyone stumped. It's still 'sitting' in my subscriptions list and one day I hope to crack it.

Best wishes, stay safe,

Bob

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

I've renotated post #6's image to try and show more clearly how my solution works, including that M must lie on LC.

Luckily for me, BM is 5, or I'd be well and truly stuck!

I wonder what maths level this puzzle is aimed at.

*Last edited by phrontister (2020-04-12 17:29:56)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,930

Why it is dangerous to assume things because they 'look right'.

Here we have any triangle ABC

Bisect A and draw the perpendicular bisector of BC; these lines meeting at O.

Construct OF, with F on AB and OFB = 90.

Construct OE similarly.

In triangles AFO and AEO, one angle is equal, another is 90, and the side AO is common, so they are congruent. Therefore, AF = AE, and FO = EO

In triangles OFB and OEC, OB = OC as triangle OBC is isosceles, angle E = angle F = 90, FO = EO from above. So they are congruent and FB = EC.

AB = AF + FB = AE + EC = AC. Therefore ABC is isosceles.

But ABC was any triangle, so all triangles are isosceles!

My thanks to Wiki for this proof.

Bob

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

Hi Bob;

The Escher Waterfall drawing (see image) uses certain visual techniques to show water flowing uphill from the wheel to the upper aqueduct...which water can't do.

Likewise, the Wiki drawing uses incorrect visual sizes of A's intersection angles and BC's intersection lengths, and shows AOD as a crooked line...which it can't be.

Conclusion: Escher's and Wiki's drawings are illusions, each representing something that is untrue as being true.

Conversely, my drawing in post #12 shows that LC is actually a straight line, with M lying on it.

*Last edited by phrontister (2020-04-16 19:56:30)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

phrontister wrote:

Luckily for me, BM is 5, or I'd be well and truly stuck!

I wondered if I could find something...and I think I might've.

Using Geogebra, I drew parallelogram LKCM (a fixed-dimension shape).

Rotating LKCM around hinge point L changed AB and BC by equal amounts, and I spotted the following formula that finds AB for a range of BMs:

AB = KL + BM

This formula is simply from observation, and I don't know any other means of arriving at it.

Here are images of solutions for 3 sample BMs:

And here is a Geogebra animation of solutions for the range BM = 5.9 to 3.9 (accuracy to 1 decimal place).

*Last edited by phrontister (2020-04-20 18:19:26)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,930

hi phrontistery,

I'm confused about this:

I drew parallelogram LKCM (a fixed-dimension shape).

but can nevertheless offer this:

edit: This seems to just go round in circles, but, if I understood the above, I think I can prove your result.

Therefore

Using Bob Creamer's values for sin and cos x we get

So it looks like you've found another way to compute AB.

Bob

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

Hi Bob;

bob bundy wrote:

I'm confused about this:

I drew parallelogram LKCM (a fixed-dimension shape).

Sorry, but for brevity's sake I omitted some steps and comments...which are:

1. I drew LM, which from the 3:4:5 triangle rule on ΔLKM = 5.

2. Then I drew KC = 5 (from being parallel to LM, and KL being parallel to BM).

3. CM = 3 (from the 3:4:5 triangle rule on ΔCMK).

4. As the opposite sides of LKCM are parallel, LKCM is a parallelogram.

5. Also, LKCM's 4 angles and side lengths are constant because of the right angles at K and M, and so LKCM retains its 'fixed-dimension shape' while it rotates around its hinge point L...a critical feature enabling me to draw the conclusion I did about AB and BC changing in length by equal amounts.

I'll have to look at your proof tomorrow coz I'm off to bed now, but thanks for your work!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,653

Hi Bob;

Thanks, that works perfectly well...and nice proof!

The formula solves AB for a range of BMs (can be non-integer) without using the 3:4:5 triangle rule. LK is always 3.

*Last edited by phrontister (2020-04-24 21:43:37)*

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