geometry questions (not looking for a detailed answers but sketchy proof or guidance can suffice)?

I now have a resolution on question 1. Starting at the top, down to "Note"; then continuing from, "OK".

1. WLOG, the length of AB can be taken as 1; this can be scaled to any desired length afterward.

Starting from the initial position, with

A=(0,0); B=(0,1); Let the initial position of C(0)=(p,q), with p,q≥0

a later position is:

A=(s,0); B=(0,c), with s²+c²=1; Let θ=sin˜¹s; then cosθ=c; θ's range is [0,π/2]

C's position is its initial position, rotated leftward by θ, then translated rightward by s:

C(θ) = Rot(-θ)C(0) + (s,0) = (cp-sq+s, sp+cq)

Note:

With this parameterization, it should be possible to put it into conic section form, and I'm still working on that. The trick is getting the s's and c's to disappear.

My idea here is to portray what the "slide" transformation does to the whole plane, since the entire first quadrant, anyway, is where point C can start out from. Part of the answer is that it is a rigid-body transformation in R².

Taking s = sinθ, c = cosθ, if you represent C(0): (p,q) in the basis:

u^₁ = (1,0); u^₂ = (0,1)

C(0) = pu^₁ + qu^₂

and that basis then moves (rotates) along with the "slide":

u^₁(θ) = (c,s); u^₂(θ) = (-s,c)

while the origin moves (translates) as

O(θ) = (s,0), then

C(θ) = pu^₁(θ) + qu^₂(θ) + O(θ)

= (cp-sq+s, sp+cq)

Try some examples:

I) C(0) = (p,q) = (1,1)

C(θ) = (c, s+c)

y = s+c = c+√(1-c²) = x+√(1-x²)

(y-x)² = 1 - x²

2x² - 2xy + y² - 1 = 0, obviously an ellipse

II) C(0) = (p,q) = (1,0)

C(θ) = (s+c, s)

x = s+c = s+√(1-s²) = y+√(1-y²)

x² - 2xy + 2y² - 1 = 0, obviously another ellipse

III) C(0) = (p,q) = (0,1) = B

C(θ) = (0, c)

x = 0, a straight line, confirming what we've already required of B

IV) C(0) = (p,q) = (½,½)

C(θ) = ½(s+c, s+c)

y = x, a straight line

OK, this is my Eureka! -- I've got it!

If you put C(0) = (p,q) into polar coordinates, (r,ϕ), so that

p = r cosϕ; q = r sinϕ, then

C(θ) = (cp-sq+s, sp+cq) = (r cos(θ+ϕ) + sinθ, r sin(θ+ϕ))

... [let θ' = θ+ϕ]

= (r cosθ' + sin(θ'-ϕ), r sinθ')

= (r cosθ' + cosϕ sinθ' - sinϕ cosθ', r sinθ')

rx = (r - sinϕ)√(r²-y²) + cosϕ y

Ax² + Bxy + Cy² + F = 0, where

A = r²

B = -2r cosϕ

C = cos²ϕ + (r - sinϕ)²

F = -r²(r - sinϕ)²

(rx - cosϕ y)² + (r - sinϕ)²y² = r²(r - sinϕ)²

So it IS an ellipse in general; with some straight-line special cases.

2. In polar coordinates, a unit circle, centered at (0,1), tangent to the x-axis, is

r = 2 sinθ

If you put one of the points at the origin, then the angles (θ coordinate) to those n-1 other points are

θ[j]=jπ/n, j=[1,n-1]

So the chord lengths are

2 sinθ[j] = 2 sin(jπ/n)

and their product is

2^(n-1) ∏[j=1,n-1] sin(jπ/n)

Trying out the cases of n = 2,3,4,6,8, I get products of 2,3,4,6,8, respectively. Why do I smell a rat?

Symmetry will pair up equal factors from both ends of that product, with one lone diameter in the center when n is even.

But that line of attack doesn't seem fruitful. And it's very inelegant to split into cases, a result that is so obviously seamless.

1)

I deleted my grossly incorrect attempt to solve this question and will try it again later.

It is true that I mistook a and b as constants which was a gross error. May be that Falzoon is right in concluding it to be an ellipse which may be a circle as a special case of an equilateral triangle, but I need to work out the correct solution.

Edit:

My partial solution of 1) is as under.

Lengths of sides of triangle ABC are constant.

Let A = (m, 0), B = (0, n) and C = (x, y)

=>

x^2 + (y - n)^2 = a^2 ... ( 1 )

(x - m)^2 + y^2 = b^2 ... ( 2 ) and

m^2 + n^2 = c^2 ... ( 3 )

Eliminating the variables m and n from the three equations should give the equation of the locus.

I could not do this last step.

Edit:

I have tried to eliminate m and n as under:

From eqn. ( 1 ) and ( 2 ),

m = x ± √(b^2 - y^2) and

n = y ± √(a^2 - x^2)

Plugging in eqn. ( 3 ),

[x ± √(b^2 - y^2)]^2 + [y ± √(a^2 - x^2)]^2 = c^2

=> x^2 + b^2 - y^2 ± 2x√(b^2 - y^2) + y^2 + a^2 - x^2 ± 2y√(a^2 - x^2) = c^2

=> 2x√(b^2 - y^2) + 2y√(a^2 - x^2) = ± (c^2 - a^2 - b^2)

=> 4x^2 (b^2 - y^2) + 4y^2 (a^2 - x^2) + 8xy √[(b^2 - y^2)(a^2 - x^2)] = (c^2 - a^2 - b^2)^2

=> b^2x^2 + a^2y^2 - 2x^2y^2 + 2xy√[(b^2 - y^2)(a^2 - x^2)] = k (a constant).

Edit:

Falzoon's prediction of the locus being ellipse can be inferred as under.

We can think of the vertex C of the triangle ABC to be very close to the line AB so that C is almost on the line AB. In this case, the locus of C is an ellipse can be proved easily. From this, it can be inferred that the locus is elliptical or nearly so.

Edit:

I tried your suggestion for triangle, square and hexagon and interestingly found the integer answers which were the same as the number of sides, 3 for triangle, 4 for square and 6 for hexagon. Interesting!! I shall try this problem sometime today when I get sufficient time.

Edit:

I checked using Wolfram Alpha for n = 3 to 11 that the answer equals the number of sides of the polygon, but could not solve a general proof for any value of n.

Hello!

I am not totally confident that I have 'pictured' the problem of the locus problem (1) you set.

So i have created dynamic geogebra document (This amazing software is FREE!!) which is my understanding of what you mean by 'ladder' AB sliding to the 'floor'

In my doc i've fixed point C by specifying angles ABC and BAC.

For this doc i've using point B as the starting point of the construction so that point A on the x axis is 'defined' wrt to B by being √( 2^2 + 3^2) away hence when B is at its starting point of (0,2) A is at (3,0)

Clearly the document confirms what you would like confirmed analytically , namely that the locus of C IS A STRAIGHT LINE.

Please confirm that the set up of the sliding ladder is correct. Then i'll be willing have a go at finding the required analytic solution.

EDIT :

=====

All hail Falzoon! :)

Bigger diagram constructed === SEE VERSION 2 === and it helped by taking on board the idea that B moves from (0,0) up the y axis

So

what i offer now is diagramatic proof that the locus is NOT A STRAIGHT LINE but looks suspiciously like an ellipse.

But proving that analytically will be no easy matter for me.

Wish me luck .... :)

EDIT:

====

Progress update: As feared, depressingly slow!

Insight just mastered that i can offer:

As point A moves towards origin at say 2 units per time interval , then point B at the other end of the ladder does NOT climb up the 'wall' of the y axis at the same rate!

Eg as A moves 2 units from (8,0) to (6,0) then we find that B moves more rapidly up , by >5units ,

from (0,0) to ( 0,√28 )

@Mathematisian ; where is Daftary's answer that you refer to?

EDIT:

====

@kelcey : cf your post of the dynamic geogebra doc

SUPERWOW ! I abase myself before a master of GEOGEBRA4 :)

I hope others begin to appreciate what a fantastic FREE resource GEOGEBRA can be as an aid to visualizing a problem.

As for myself sorting out my pc so that GEOGEBRA4 works on it ( i'm stuck on 3.2) has just been bumped up my to-do list!

@Fred. I'm sure you have the analytical solution to 1)

unfortunately my maths is rather too rusty to easily follow your argument

basically can't follow

"

My idea here is to portray what the "slide" transformation does to the whole plane, since the entire first quadrant, anyway, is where point C can start out from. Part of the answer is that it is a rigid-body transformation in R².

Taking s = sinθ, c = cosθ, if you represent C(0): (p,q) in the basis:

u^₁ = (1,0); u^₂ = (0,1)

C(0) = pu^₁ + qu^₂

and that basis then moves (rotates) along with the "slide":

u^₁(θ) = (c,s); u^₂(θ) = (-s,c)

while the origin moves (translates) as

O(θ) = (s,0), then

C(θ) = pu^₁(θ) + qu^₂(θ) + O(θ)

= (cp-sq+s, sp+cq)

"

can you offer an online resource that would enable me to brush up on the maths you are using in the quotes?

1. I'm not at all proficient in loci, but there are plenty of others, who no doubt, will

come to my aid with the appropriate equations. All I could do was test it out in

practical terms, for which I used an equilateral triangle, and found that the locus

appears to be part of an ellipse. It's definitely not a straight line or a circle.

2. By drawing radii and using the sine rule, I get :

2*[sin(36º)/sin(72º) + sin(72º)/sin(54º) + sin(108º)/sin(36º) + sin(144º)/sin(18º)] + 2

which can be simplified (not easily) to : 2[1 + √5 + √(5 + 2√5)].

3. I don't think I'll get around to tackling this one.

I'll star this question because I'm interested in furthering my knowledge on loci,

in particular.

Edit: I didn't read Q.2 correctly, so added instead of multiplied,

but I won't be able to get back to it until later.

Edit: For question 2, if I've done it correctly this time, I get :

[sin^2(36º)/sin^2(72º)] * [sin^2(72º)/sin^2(54º)] * [sin^2(108º)/sin^2(36º)]

* [sin^2(144º)/sin^2(18º)] * 2

which simplifies to 2*sin^2(108º)*sin^2(144º) / [sin^2(54º)*sin^2(18º)]

which further simplifies to 10.

EDIT for Q.1:

'Tis better that I deleted the rubbishy speculation. This is all I'm sure of, so far -

For an equilateral triangle, as point A moves along the X-axis, with θ = angle CAX,

then all (x, y) coordinates of C on the locus are of the form :

( cos[θ] + cos[2π/3 - θ], sin[θ] ), where π/6 ≤ θ ≤ 2π/3.

I don't yet know whether or not those points lie on an ellipse.

EDIT2 for Q.1: Little bits at a time. Well, they do seem to fit the ellipse,

4x^2 + 4y^2 - (4√3)xy - 1 = 0, where the semi-major axis is a = (√3 + 1)/2

and the semi-minor axis is b = (√3 - 1)/2.

Eccentricity, e = √(4√3 - 6), focal distance = 3^(1/4), centre = (0, 0).

http://www.wolframalpha.com/input/?i=4x%5E2+%2B+4y...

If this ellipse is rotated by -45º, the equation is x^2/a^2 + y^2/b^2 = 1, where

a and b are, as above. Here are both ellipses shown together -

http://www.wolframalpha.com/input/?i=4x%5E2+%2B+4y...

1. As AB slides along the axes the mid-point M moves on the quarter circle centre

O and radius OM since OM=MB=MA.

The line MC is fixed in length and the angle OMB and BMC are fixed and so

therefore is the angle OMC. This means that the line OC is of fixed length. The locus of C

is part of a circle centre O and radius OC.

2. I can only see the complex number method.

3. Use the parametric form x=t^2, y=2t (letting a=1) then the area of triangle

ABC is (1/2) |Ax(By-Cy)+Bx(Cy-Ay)+Cx(Ay-By)| where Ax is the x-coordinate of A etc.

This gives the area of triangle T1T2T3 =|(t1t2(t1-t2)+t2t3(t2-t3)+t3t1(t3-t1)|

The tangents at T1 and T2 meet at (t1t2, t1+t2) and similarly for the other pairs

so you can use the formula in line 2 to get the required area.

VERY INTERESTING! Thanks for posting the Q!

Recap:

"➊ ∆ABC lies in the first quadrant such:

• that A and B lies on the X and Y axis respectively

• C lies on the non origin side of AB.

What (geometric figure) is the locus of the point C when AB slides (like a ladder) with end points on both the axes:

• A moves from (0, 0) to (|AB|, 0) on the coordinate x-axis

• B moving from (0, |AB|) to (0, 0) on the coordinate y-axis

Appears that locus is a straight line but was looking for a analytical proof...."

I will add my 2¢ worth wrt # ➊ ...

Similar to my friend, "slowjerry", I too have prepared a dynamic file (using "FREE" GeoGrebra) that (once created) allows one to alter:

• whether or not D, a point on AB, is the midpoint of AB ... or not

• the height of the ∆ABC, |CD| , from 0 to whatever value you like

... and view the resulting locus of point C.

To get a better 'feel' for what would happen in all four quadrants, I asked GeoGebra to create rotation images of the quad 1 ∆ABC, to simulate what would happen to the locus of C as the ∆ continued rotating beyond quad 1. By doing so, it confirms one's gut 'suspicions' that ALL the loci are ellipses with centres at the origin, and with special cases being: the circle; and the 'degenerate' ellipse ie a straight line.

While not a 'proof' as such, the ability to 'experiment' with the controls is:

a) fun

b) informative

c) leads one to certain instructive perceptions

The picture of the locus when:

• point D is the midpoint; |DC| = 0? ... a circle:

http://sogacity.com/misc/images/screengcg.png

• point D is the midpoint; |DC| > 0? ... an ellipse that has the major axis tilted 45º angle to the x-axis:

http://sogacity.com/misc/images/screenltl.png

• point D is NOT the midpoint; |DC| = 0? ... an ellipse with major & minor axes on the x-&y-axis:

http://sogacity.com/misc/images/screensis.png

• point D is NOT the midpoint; |DC| > 0? ... an ellipse that has the major axis tilted at OTHER THAN than a 45º angle to the x-axis

http://sogacity.com/misc/images/screenwuw.png

Determining the equations of the ellipses is another 'kettle of fish' that I will endeavour to determine (using GeoGegra).

Hope fully I can have something before the Q expires.

Hope this helps! Cheers! :)

.

===== EDIT #1 ====

Still 'playing around' with my creation.

Here is a pic showing multiple superimposed loci of C with:

• CONSTANT location of point D on AB

• incrementally increasing |DC| beyond 0 to ... whatever

http://sogacity.com/misc/images/screenxlx.png

Observations:

• angle of tilt continues to increase with the length of |DC|

• the major vertices of the ellipses created SEEM to be following a hyperbolic path

• eccentricity seemed near a minimum when |DC| ≈ ½|AB| ie when DC = 150 (my |AB| is 300)

Here is a pic showing multiple superimposed loci of C with:

• CONSTANT |DC| .

• incrementally changing the location of point D on AB from:

••• D coincident with A

••• ➞ midpoint D

••• ➞ to D coincident with B

http://sogacity.com/misc/images/screenqyq.png

Observations:

• angle of tilt (wrt x-axis) continues to increase with the location of D (as D approaches B from A)

• the major vertices of the ellipses created SEEM to be following a rectangular hyperbolic path

• eccentricity appears to stay constant

(Will be playing for a bit ... waiting for a Eureka moment that sometimes occur when playing')

Will return if I discover anything else of interest. :•)

.

==== EDIT #2 ====

BTW:

If anyone is interested in playing with this file themselves, here is a link to the HTML version of the GeoGebra file that I created (from which the screen shots were taken above):

http://dl.dropbox.com/u/62787550/LocusOfPointC.htm...

.

===== EDIT #3 =====

I had said above ... :

"

Determining the equations of the ellipses is another 'kettle of fish' that I will endeavour to determine (using GeoGegra).

Hope fully I can have something before the Q expires.

"

... and I HAVE definitely had some "Eureka Moments!" ... BUT ...

... I will likely NOT be able to complete a well-presented set of thoughts before the Q expires as we leave tomorrow (21st) for a late-winter holiday to sunnier climes. I will however add my findings after our return (≈ March 10th) as an email to whomever desires to receive what I (using GeoGebra) will have come up with. If interested, email me with your desire to see what I have come up with.

.

Regards to all

Neil

.

[ 3 ]

Let P(t₁), Q(t₂) and R(t₃) be on the parabola y² = 4ax.

Then : P ≡ ( at₁², 2at₁ ), Q ≡ ( at₂² , 2at₂ ), R ≡ ( at₃², 2at₃ ).

If the tangents at (i) Q and R meet in P', (ii) R and P meet in Q'

and (iii) P and Q in R', then :

P' ≡ ( at₂t₃, a(t₂+t₃) ), Q' ≡ ( at₃t₁, a(t₃+t₁) ), R' ≡ ( at₁t₂, a(t₁+t₂) ).

Now, we can show that

area of ΔPQR = a² (t₁-t₂)·(t₂-t₃)·(t₃-t₁)

and

area of ΔP'Q'R' = (1/2)· a² (t₁-t₂)·(t₂-t₃)·(t₃-t₁)

so that :

area of ΔPQR = 2· ( area of ΔP'Q'R' ). .................. Q.E.D.

_____________________________________