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**Prove this**

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ABCD is a square

PQ = QC = DR

Prove that: angle QPR = 45 degrees

Caution: I am not sure if I have drawn the following diagram perfectly or correctly

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Hi Agnishom;

Are you sure it is DR?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,394

hi Agnishom,

Hhhmm. as given, that diagram is impossible.

DR > DC > QC, so DR = QC is impossible.

Bob

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

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How about it being BR?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,394

I think that might be possible. I'll get to work on it.

LATER I've used Sketchpad to make an accurate construction and QPR = 46.13

So that's not it I'm afraid.

EVEN LATER:

I've made a construction for the square, the property PQ = QC and QPR = 45.

When I use Sketchpad to measure angles and distances the software puts in letter labels automatically, so I've ended up with F,G and H rather than P, Q and R. But the diagram has the right features apart from that. I cannot find a third length equal to PQ and QC.

Bob

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Hi;

But if there are lots of points that have the condition

QC = PQ = BR each having a different angle QPR how can you prove it is 45 degrees?

Shouldn't this question be find the points P, Q and R where angle QPR is 45 degrees?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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I am sorry

I will try to find out the original question from whom I got it once this formative assesment is over

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Hi;

I have a solution that finds the points you need. My solution shows that PRB is a right angle which is not shown in your drawing or given.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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You seem correct.

How about the proof?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,394

hi Agnishom and bobbym

Here's two new diagrams.

I constructed the square ABCD.

I put point G somewhere on AD

I made an isosceles triangle FGC so that FG = FC

I found the centre of the square and rotated point F by 270 degrees around this point to fix H.

That makes BH = FG = FC

I constructed a line, perpendicular to GF at G and bisected the angle to make FGJ = 45.

As you can see in the first diagram J and H are different points.

But, this method of construction allows me to move G along the line. My plan was to find the position where J and H coincide.

Because of the size of the points, it's hard to be exact about this (Euclid had dimensionless points but they cost extra )

So I measured GHB and GJB and tried to make them equal. The second diagram shows my best attempt.

So the 90 case seems to be a special case and the only one where FGH = 45.

(Not counting G at A or B)

Your original diagram does make it look like there's a right angle there, but, in geometry, it is dangerous assuming things just because they look like it.

Later I'll have a go at the proof when I assume the angle is 90.

Bob

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,394

I think you can do this yourself.

Hint: draw the line FK so that K is on GH and FK is perpendicular to GH.

Bob

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Good Heavens! What a confusion!

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

You seem correct.

How about the proof?

Actually, the diagram is the proof. A proof without words. It uses coordinate geometry and trigonometry to get every side and every angle. To have that drawing one must find that angle PRB is a right angle which I said earlier.

The diagram does not mention the right angle so finding the solution, where P, Q and R is all that is necessary. To prove that angle is 45 degrees was not possible without the given of angle PRB as 90 degrees.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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