Math Question

[1fast1]

I'm thinking of playing around with vertical and horizontal bisectors -

wondering if that would help at all

 

I also notice that the angles are "fun"

 

(i.e 15 / 75 / 150 / 60) - all are multiples of 15 - and wondering if this has some sort of impact

Good observation, and yes, all the involved angles are multiples of 15 degrees. If you were solving with trigonometry it would probably be more impactful, but it's just a nice geometric observation.

 

I'll just say you're correct that extra construction is needed to resolve the problem so you're on the right general trail. Would bisectors really be that helpful though? I'm not sure cos I didn't do that personally.

[Darryn]

Good observation, and yes, all the involved angles are multiples of 15 degrees. If you were solving with trigonometry it would probably be more impactful, but it's just a nice geometric observation.

 

I'll just say you're correct that extra construction is needed to resolve the problem so you're on the right general trail. Would bisectors really be that helpful though? I'm not sure cos I didn't do that personally.

Dude -

I was *this* close to failing calc at my last pre-U year, and haven't done math since -

My arithmetic is fine, but much beyond that I am floundering

[Enye]

Dude -

I was *this* close to failing calc at my last pre-U year, and haven't done math since -

My arithmetic is fine, but much beyond that I am floundering

This type of question just keep drawing different lines through point m then find the values of all the angles you can find then conclude it is equilateral

 

Confuse the person marking that you know but actually don't

 

😂

[1fast1]

Boarding now. Will post solution later.

[Angcheek]

Boarding now. Will post solution later.

 

hmmm I still need to have 1 assumption.

 

ie. for DCM to be equ , all sides must be equal length. DM = DC which also = AD

 

if length of AD = DM , angel DAM = AMD . (since we know DAM = 75)

therefore ADM = (180-75-75) = 30

so CDM = 90-30 = 60 (applied to both side of sq) .

 

hence DMC also 60.

[Invigorated]

Geometrically, these 2 triangles are only possible if PAC and CPB are both quadrants.

 

PM, PC and CM are all radius of the quadrants but this is pure observation. It has to be an equilateral triangle but this isn't with mathematical proofs.

 

I'm not able to calculate why it must be 15 degrees with the exception of using pythagoras and trigo so guess I'm not quite there.

Edited October 9, 2015 by Invigorated

[ST69]

How do you know (prove) the side triangle is isosceles? You're making a large assumption there. Only the base of triangle DMC can be assumed to be the side length of the square. The other two sides are not given.

 

If you assume that "side" triangle is isosceles then you don't even need all that. The other two lengths of DMC are now also equal to its base so it's equilateral. No need to even consider angles.

 

But the point is: you have to show the side triangle is isosceles in the first place.

 

Oh...not assuming, i thot that was obvious...Ok, ok...will attempt to put that in words since its been > 30yrs since I last solve such problem.

 

step by step...use tangent beta = opp / adj

 

1. draw horizontal line across M perpendicular to line BC at point E, where ME=x

2. tan MCE = ME / CE ... CE= x / tan MCE

3. tan MBC = tan 75 = ME / EB ... EB= x / tan 75

4. CE+EB = 2x,

5. work out the 3 equations, solve MCE = 30, hence DMC = 2x30 = 60 (since triangle DMA=CMA)

therefore,DMC=equilateral triangle.

 

 

Edited October 9, 2015 by ST69

[Darryn]

Answer is available online.

 

It uses properties of triangles I dun unnerstand

 

But is logical.

[1fast1]

 

Oh...not assuming, i thot that was obvious...Ok, ok...will attempt to put that in words since its been > 30yrs since I last solve such problem.

 

step by step...use tangent beta = opp / adj

 

1. draw horizontal line across M perpendicular to line BC at point E, where ME=x

2. tan MCE = ME / CE ... CE= x / tan MCE

3. tan MBC = tan 75 = ME / EB ... EB= x / tan 75

4. CE+EB = 2x,

5. work out the 3 equations, solve MCE = 30, hence DMC = 2x30 = 60 (since triangle DMA=CMA)

therefore,DMC=equilateral triangle.

 

 

Right at the start, I said "no trigonometry". Trig makes it quite trivial especially if a calculator is allowed.

 

Geometry only. So I'm sorry but your solution is not what is being sought.

[1fast1]

Will check back here in a few hours. If no one has posted a valid solution using only basic geometry (no trig!!) I will post images of the solution I scribbled down while in Frankfurt.

[ST69]

Right at the start, I said "no trigonometry". Trig makes it quite trivial especially if a calculator is allowed.

 

Geometry only. So I'm sorry but your solution is not what is being sought.

 

Oh...you mean by drawing..sori I no academic.. ....here goes... basically, ratio, angles & length of square.

 

Many ways to skin a cat... (Oops .. no offence to cat lover... )

Edited October 10, 2015 by ST69

[Invigorated]

I was thinking about the 15 degrees. If we draw a similar triangle, we will get an equilateral triangle and with a bisector of the equilateral triangle, we will gather that DA is equal to DM and hence DMC must be equilateral.

[ST69]

another way to skin it...

 

I was thinking about the 15 degrees. If we draw a similar triangle, we will get an equilateral triangle and with a bisector of the equilateral triangle, we will gather that DA is equal to DM and hence DMC must be equilateral.

 

[1fast1]

 

Oh...you mean by drawing..sori I no academic.. ....here goes... basically, ratio, angles & length of square.

 

Many ways to skin a cat... (Oops .. no offence to cat lover... )

No offence but your method is not clear and could use a lot more explanation. In any case it really looks like you're still using trigonometric properties without calling them by the usual names. Edited October 10, 2015 by Turboflat4

[1fast1]

I was thinking about the 15 degrees. If we draw a similar triangle, we will get an equilateral triangle and with a bisector of the equilateral triangle, we will gather that DA is equal to DM and hence DMC must be equilateral.

And we have a winner. This construction is what I arrived at as well. Basic geometry after that. Well done!

Bit irrelevant now as @Invigorated already got it but since I promised here are my hotel scrawls. A tad lengthier than the congruence argument used by Invigorated but otherwise the same idea.

Edited October 10, 2015 by Turboflat4

[1fast1]

What I find most interesting about this problem is how much more difficult it is to go in one direction than another. If you're given the big triangle is equilateral it's almost trivial to show that the little triangle has base angle 15 degrees. But going the other way requires real inspiration.

[ST69]

No offence but your method is not clear and could use a lot more explanation. In any case it really looks like you're still using trigonometric properties without calling them by the usual names.

 

None taken...Thanks for the good distraction & exercise for my brain cells...

[Thaiyotakamli]

Calling math gurus for the answer