Cool problem of the day! (1 Viewer)

[Carrotsticks]

Figured this would be a good way for some students to practise problem solving skills.

I'll release a question every day or so and if the problem requires a technique/formula outside the syllabus, I will tell you what that formula is.

Consider the following diagram.

AC is perpendicular to BE and D is the midpoint of BE.

Show that x = z if and only if Angle BAE is 90 degrees.

 

[barbernator]

Show that x = z if and only if Angle BAE is 90 degrees.

I solved the question, yet i'm not sure if my proof is the correct way to prove it.

To start the proof, would it be ok to say "Let angle BAE be a right angle" and go from there?

 

[Carrotsticks]

I solved the question, yet i'm not sure if my proof is the correct way to prove it.

To start the proof, would it be ok to say "Let angle BAE be a right angle" and go from there?

That would prove one side of the iff, but you would have to do the other side too.

 

[deswa1]

Would you do it first by assuming BAE isn't 90 degrees and showing that x cannot equal z and then showing that they do equal when BAE is 90? I haven't tried the problem yet but I'll have a go.

 

[funnytomato]

I solved the question, yet i'm not sure if my proof is the correct way to prove it.

To start the proof, would it be ok to say "Let angle BAE be a right angle" and go from there?

This might be somewhat useful:

as Carrot said, 'if and only if' implies that there are 2 separate(that's how it's spelt, right?) proofs
[which may or may not use the same kind of technique]
namely 'if' and 'only if'

for this particular question: 'Show that x = z if and only if Angle BAE is 90 degrees.'

if : if angle BAE=90° , prove angle x = angle z
only if: if angle x = angle z , prove that angle BAE MUST BE 90°, hence ' ONLY if '

 

[funnytomato]

Would you do it first by assuming BAE isn't 90 degrees and showing that x cannot equal z and then showing that they do equal when BAE is 90? I haven't tried the problem yet but I'll have a go.

i think these would be sufficient

 

[Carrotsticks]

Not entirely convinced that is sufficient. You are using a proof by contradiction, but it doesn't necessarily mean that doing so means you cover both sides of the argument in 1 go.

 

[funnytomato]

@Carrotsticks, this isn't really out of the syllabus , or is it?
also 'if' and 'only if' can be proved basically using the same technique , right?

 

[kazemagic]

O_O can a year 10 do this?

 

[Carrotsticks]

@Carrotsticks, this isn't really out of the syllabus , or is it?
also 'if' and 'only if' can be proved basically using the same technique , right?

This particular question is in-syllabus but later on, I will be churning out some Number Theory and various 'Olympiad-type' questions, which are most certainly not in the syllabus!

iff can also be proved through contradiction (or 'controdiction' according to Florica Cirstea! Drives me crazy...) but then that means you would have to assume the opposite, then prove both sides are true anyway, so may as well do it straight, though I can imagine there would be some problems where "contradiction, then iff" is more appropriate than "iff" directly.

 

[Sanical]

suppose ABC lies on a circle

if x + y + z = 90 degrees, then AD = DE as equal radii, thus x = z as ADE becomes an isosceles triangle

if angles x + y + z does not equal to 90 degrees, BDE would not be the diameter implying DA does not equal to DE

unsure how to prove that last part although i know its true....

Haha, very nice .

 

[Carrotsticks]

suppose ABC lies on a circle

if x + y + z = 90 degrees, then AD = DE as equal radii, thus x = z as ADE becomes an isosceles triangle.

I don't get how you deduced the last part based purely on what you've given us there. If ADE is an isosceles triangle, then Angle DAE = Angle DEA = z.

Care to tell us how you deduced x=z just from that?

 

[Sanical]

I don't get how you deduced the last part based purely on what you've given us there. If ADE is an isosceles triangle, then Angle DAE = Angle DEA = z.

Care to tell us how you deduced x=z just from that?

Yeah, I didn't get how he deduced that. It still took me a few lines to show x=z if it were a circle.

 

[qwerty44]

I don't get how you deduced the last part based purely on what you've given us there. If ADE is an isosceles triangle, then Angle DAE = Angle DEA = z.

Care to tell us how you deduced x=z just from that?

Wouldn't it be because AD=BD (equal radii of circle)

Therefore Triangle ABD is isosceles

So Angle ABD=90-x (angle sum of right triangle ABC)

and Angle AEB 90-(90-x)=x (angle sum of right triangle ABE)

But Angle BEA=Angle DAE (as shown by that other guy)

Therefore z=x

???

 

[qwerty44]

Carrotsticks would you mind taking me through how to set out an "if and only if" proof, because I really don't know.

 

[Carrotsticks]

InTrix has got it, nice.

I will post up a solution for this question later on. For the mean time, here is another one, but I warn it is pretty difficult:

I actually posted it up in the Uni Maths section earlier this week and would sorta come under the topic "Geometric probability".

http://community.boredofstudies.org/showthread.php?t=282730

Consider a circle of any radius (does not matter). We select two randomly and independently selected points on the circumference and as a result, we have a random chord. If R random chords are drawn, show that the probability that NONE of them intersect is....

 

[Drongoski]

Carrotsticks would you mind taking me through how to set out an "if and only if" proof, because I really don't know.

Say you want to show that triangle ABC is right-angled if and only if the sum of the squares of the 2 shorter sides equals the square of the longest side.



Then one standard approach is to show that the 2nd proposition follows from the 1st and then show the 1st proposition follows from the 2nd. Another way of saying this is: the LHS implies the RHS and the RHS implies the LHS as well. Thus for the above example we need to show:



and

 

[qwerty44]

Wouldn't it be because AD=BD (equal radii of circle)

Therefore Triangle ABD is isosceles

So Angle ABD=90-x (angle sum of right triangle ABC)

and Angle AEB 90-(90-x)=x (angle sum of right triangle ABE)

But Angle BEA=Angle DAE (as shown by that other guy)

Therefore z=x

???

Say you want to show that triangle ABC is right-angled if and only if the sum of the squares of the 2 shorter sides equals the square of the longest side.



Then one standard approach is to show that the 2nd proposition follows from the 1st and then show the 1st proposition follows from the 2nd. Another way of saying this is: the LHS implies the RHS and the RHS implies the LHS as well. Thus for the above example:



and

What about in terms of this question? (the first one in the thread)

 

[Nooblet94]

What about in terms of this question? (the first one in the thread)

You would show x=z implies BAE is a right angle and then (seperately) show that BAE being a right angle implies x=z

 

[Fus Ro Dah]

http://community.boredofstudies.org/showthread.php?t=282730

There are (r-1)!! ways of drawing r chords.

We will utilise the Catalan Numbers and bracket strings to calculate the probability.

Define a 'good' bracket string to be closed ie: (()()) or ()()() where B_left = B_right, where these bracket strings represent chords.

A 'bad' bracket string is where B_left =/= B_right.

The combinatoric definition, and I say this because there is a product definition too, of the Catalan Number is:



This is the number of ways we can have r 'good' strings.

Therefore:



Where we simply turned the double factorial into single factorial notation.