Brain Teaser

[SydneyRoverP6B]

Hello Richard,

Your triangle - circle problem certainly works the brain.!! Lots more time to be spend there.

With your clue,...quattro wrote,...

Clue - may not be similar, but certainly congruent

You meant the other way,...congruent triangles are always similar, but the converse is not always true.

Ron.

 

[SydneyRoverP6B]

Hello Richard,

I am definitely not happy with my answer above, so I am trying a different approach. There are so many unknowns, similar triangles, simultaneous equations,...full of unknowns.

It certainly occupies my mind,...keep coming back to it as I like to work things out.

Ron.

 

[quattro]

Its all to do with congruent triangles and finishes up with a quadratic equation 8)

Mark the centre point of the circle and draw in three radii - that'll help

Richard

 

[SydneyRoverP6B]

Hello Richard,

I have just finished it again,...great problem! Used similar triangles, derived some equations and solved for the unknown. AC is according to my calcs 17.6, but of course variations may occur due to rounding.

The 3 sides of the triangle form tangents to the circle, and the radii are orthogonal to the tangents. The diagram you posted is exactly as I have drawn except I did not put in AD. I placed a chord EF and then used similar triangles,...EDH (H is the midpoint of EF) and BDF, constructed equations and solved the length DB. Take arcsin of the ratio to obtain the angle DBF, which when doubled is the angle ABC. Then use sin of that angle along with the hyp AB to calculate AC.

How does that compare to your figures? I used quadratic equations before, but this second approach looks much better.

Ron.

 

[quattro]

How does that compare to your figures?

I haven't worked it out yet :shock:

I will have to do it later as I am at work at the moment

The way I did it those years ago, was to work out that triangle AED and AGD are congruent, i.e. identical in shape and size albeit a mirror image in the own hypoteneuse.

Also triangles EBD and FBD are congruent in the same way.

GC and FC are sides of a square and therefore equal to DG and DF which we know to be 2.5 as they are radii of a 5' circle.

So we have the theory that the diameter of the circle plus the length of the hypoteneuse is equal to the sum of the lengths of the other two sides.

I am going to call that the Cleal theory because I discovered it 8) - Yeah right LOL

I'll do the rest later when I am less busy and not attached to the phones here.

Starting from, AB (18.5) + 5 = BC +AC

 

[SydneyRoverP6B]

Hello Richard,

quattro wrote,...

So we have the theory that the diameter of the circle plus the length of the hypoteneuse is equal to the sum of the lengths of the other two sides.

I can honestly say that I had never heard of that piece of information before, not in a lecture nor from reading a text. Labelling the lengths in terms of x, r and the length of the hypotenuse, forming simple algebraic additions then confirms that theorem.

So using that wonderful piece of information the problem is now simple and can be solved in less than 5 minutes.

AB being 18.5 units and circle diameter being 5 units gives a sum of 23.5 units.

BC + AC = 23.5............(1)

From Pythogoras,...AB^2 = BC^2 + AC^2,..so

BC^2 + AC^2 = 18.5^2....(2)

Solve (1) and (2) simultaneously which gives BC = 6 and AC = 17.5.

Working with similar triangles and writing equations in order to solve for the unknown took ages and gave an answer 0.1 off the answer above. From that wonderful piece of information that you shared, there is no need to form congruent triangles or any other geometric mappings. In fact nothing needs to be drawn at all. The diameter of the circle and the length of the hypotenuse is all that is required. Form the two equations above and solve.

Wonderful, thanks Richard for a top problem....

Ron.

 

[quattro]

Hello Richard,

quattro wrote,...

So we have the theory that the diameter of the circle plus the length of the hypoteneuse is equal to the sum of the lengths of the other two sides.

I can honestly say that I had never heard of that piece of information before, not in a lecture nor from reading a text. Labelling the lengths in terms of x, r and the length of the hypotenuse, forming simple algebraic additions then confirms that theorem.

Nor me, or anyone else that I spoke to about it at the time including my Brother (A level maths teacher at the time) or any of his math buddies. That's why I am calling it the Cleal theorem 8) Cos I discovered it LOL

So using that wonderful piece of information the problem is now simple and can be solved in less than 5 minutes.

AB being 18.5 units and circle diameter being 5 units gives a sum of 23.5 units.

BC + AC = 23.5............(1)

From Pythogoras,...AB^2 = BC^2 + AC^2,..so

BC^2 + AC^2 = 18.5^2....(2)

Solve (1) and (2) simultaneously which gives BC = 6 and AC = 17.5.

Working with similar triangles and writing equations in order to solve for the unknown took ages and gave an answer 0.1 off the answer above. From that wonderful piece of information that you shared, there is no need to form congruent triangles or any other geometric mappings. In fact nothing needs to be drawn at all. The diameter of the circle and the length of the hypotenuse is all that is required. Form the two equations above and solve.

Wonderful, thanks Richard for a top problem....

Ron.

I had to form the congruent triangles to start with, because I didn't know the circle/hypoteneuse theorem.

Very well done anyway Ron. It took my Brother 3 weeks to drive himself mad, and me several hours and many trials and tribulations to get that.

 

[webmaster]

Solve (1) and (2) simultaneously which gives BC = 6 and AC = 17.5

That's what I said a couple of days back 17'6" !, of course I avoided all the maths by banging it into a cad package and reading off the lengths

 

[quattro]

Solve (1) and (2) simultaneously which gives BC = 6 and AC = 17.5

That's what I said a couple of days back 17'6" !, of course I avoided all the maths by banging it into a cad package and reading off the lengths

Ok Richard - well done to you too LOL

That is kind of cheating though - just a little.

Richard

I got an 18'6" ladder out but couldn't find a wall with a 5' pipe against it :shock:

 

[darth sidious]

Well we have a correct answer and Richard (quattro) is the first to do so. Well done Richard...

I used Calculus,...related rates and implicit differentiation to solve the problem. Time 12 mins and distance 32 root 13 km which approximates to 115.38km. Using Pythagorean triangles repeatedly to narrow down the distance will also yield the same outcome.

Hmmm....I'll have to find a tougher teaser next time.

Ron,

Here's my solution, which does match the others! I use vectors, as I find them very convenient!

For this problem, I shall assign t to have units of minutes, as opposed to seconds (again, for convenience).

At t = 0

Jet 1 is heading east at 12km/min. Let's assume that Jet 1 is at the origin at t = 0. Jet 1's displacement, which we we shall call S1

S1 = 12tî

Jet 2 is heading north at 8km/min and is 208 km away from Jet 1, at the same altitude/height. Jet 2's displacement, which we we shall call S2

S2 = 208î + 8tĵ

The displacement vector between the two is S2 - S1 = (208î + 8tĵ) – (12t î) = (208 – 12t) î + 8tĵ

The magnitude of the vector is, as always, z^2 = x^2 + y^2

x = (208-12t), and y = (8t)

Implicit Differentiation time!

2z dz/dt = 2x dx/dt + 2y dy/dt, as 2 is a common factor, we can eliminate the 2’s to get :-

z dz/dt = x dx/dt + y dy/dt.

Now, we could divide through by z, but it’s not necessary here….

z is the distance between the planes, and therefore is non-zero. (If it was zero, it would mean an air crash had occurred! :shock

For max/min values, dz/dt = 0. We want the min value, so to make sure what we work out is the min value… we work out what (d/dt)^2 {z} is

z dz/dt = [(208 – 12t) x -12] + 64t = 208t – 2496 = 208(t – 12). dz/dt is 0 at t = 12 minutes

Differentiating again gives that (d/dt)^2 {z} = 208. As (d/dt)^2 {z} > 0, this means that at t = 12 mins, the minimum value of z is found. (There is no global maximum value, as after t = 12mins, the distance keeps on increasing.)

The Euclidean distance between the planes :-

||S2 - S1|| = [(208 – 12t)^2 + (8t)^2]^0.5

Ar t = 12 mins…

t =12 min , ||S2 - S1|| = [(208 - 144)^2 + (8 x 12)^2]^0.5 = [64^2 + 96^2]^0.5

As 64 and 96 are all multiples of 32, so… 64 = 32 x 2 and 96 = 32 x 3,

32 x [2^2 + 3^2]^0.5 = 32 x [4 + 9]^0.5 = 32 x 13^0.5 = 32 x 3.60555 (5 d.p) = 115.378km (3 d.p.)

The minimum distance between the planes is 115.378km, at time t = 12 minutes.

 

[SydneyRoverP6B]

Hello Darth,

That is another good way to solve the problem, albeit a little more lengthy...

Vectors in bold font and unit vectors wearing their top hats,....excellent!

Ron.