Fox 332

Won't you come and see?
Some fly away,
some remain lonely.
-- Himalayan Fox

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Fox 325, 326, 328 - Solutions

These simple ones are all related, so their solutions are posted here together.

Fox 325 by Bob Ryden:

Fox 325 by six:

Fox 326 by Bob Ryden:

Fox 326 by Binary:


Fox 328 by César Lozada:


Fox 328 by Bob Ryden:

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Fox 326

Hard talks much, yet simple tells more.

--- Himalayan Fox

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Fox 325

We've been carried away with very hard problems recently. Let's have a simple reality check with this classic. But, please try solving this without any words :)

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Fox 302 - Solution

To see the animation below, you will need any of the following browsers:

Chrome, Firefox, Opera, Safari, or IE Explorer 9.0.

Bleaug teases the human intelligence, saying:

"I know that analytic solution to this Fox 302 is obvious (although cumbersome). Here is an alternative geometric demonstration. Use Home/PageUp/PageDown/End keys/ or mouse wheel to go though demo steps."

Do enjoy!

Fox 269 - Solution

By Yu:

0 ≤ α ≤ 60

Large Ellipse

Semi-major axis: (1/2)(√3 + cot(α/2))
Semi-minor axis: (1/2)(√3 - tan(α/2))
Area = π*major*minor=(π/2)(√3 cotα + 1).

Small Ellipse

Semi-major axis: (1/2)(tan(α/2) + √3)
Semi-minor axis: (1/2)(cot(α/2) - √3)
Area = π*major*minor = (π/2)(√3 cotα - 1).

Ratio of their areas (large : small) = (√3 cotα + 1)/(√3 cotα - 1)
= [(√3/2)cosα + (1/2)sinα] / [(√3/2)cosα - (1/2)sinα]

= sin(60+α) / sin(60-α)

60 ≤ α ≤ 90
Large Ellipse: Area = (π/2)(1 + √3 cotα)

Small Ellipse: Area = (π/2)(1 - √3 cotα)

Ratio of their areas (large : small) = (1 + √3 cotα)/(1 - √3 cotα)
= [(1/2)sinα + (√3/2)cosα] / [(1/2)sinα - (√3/2)cosα]

= sin(α+60) / sin(α-60).

An observation:
When α=60, the ratio is undefined because the area of the small ellipse is 0. The third vertex maps out a straight line.

Fox 267 / Fox 268 - Solutions

Two solutions for Fox 267 and Fox 268:

Bleaug:

Scalar values u=OU and v=OV vary according to a quadratic equation (e.g. apply cosine rule to OUV triangle)
M is a linear function of U and V, hence M coordinates obey some quadratic equation, i.e. M location is a conic. Same applies to N.
Since UV=1, these conics are finite, hence they are ellipses.
Since construction is symmetric, these ellipses are centered in O and have OUV bissectors as symmetry axes.
267: From Fox 266 we know that ON=UN=UV=1, hence N location is a circle of radius 1 centered in O.
268: The skew transform into green ellipse preserves the area. So blue/green ellipse area is twice the red circle area = 2*pi.

Yu: (267)

m(BOC) = (½) m(BAC), .: O, B and C are on the circle centred at A, .: OA = AB.
m(B'OC') = (½) m(B'AC'), .: O, B' and C' are on the circle centred at A', .: OA' = A'B'.
AB = A'B', .: OA' = OA. For other orientations refer to Fox 266.
The vertex A is always the same distance from O, so it maps out a circle of radius OA centred at O.

Yu: (268)

The two intersecting lines are y = ±(2 - sqrt(3))x.
Let A(a, (2-sqrt(3))a) and B(b, -(2-sqrt(3))b).
AB^2 = (2-sqrt(3))^2 (a+b)^2 + (a-b)^2 = 1.
By matrix methods, vertex C has x=(sqrt(3)-1)(a+b) and y=-(sqrt(3)-1)(a-b).
.: x^2/(sqrt(3)+1)^2 + y^2/(sqrt(3)-1)^2 = 1, an ellipse.
Area of ellipse = pi (sqrt(3)+1)(sqrt(3)-1) = 2pi

Fox 266 - Solutions

Solutions of Fox 266 by Joe, Bleaug, Yu, and Giannno are similar. There is a different geometric solution by Binary Descartes, we should add it in the following days.

Bleaug:
No words needed. (Hey y not? say somethin' :)

Yu:Given phi=30°, then m(BOC) = 150° which is 1/2 of reflex m(BAC).
.:B, O and C are on the same circle centred at A. Hence AB = AO.

Giannno: Extend CA such a way that AD=AC. Then m(DBC)=90° (since BA=DC/2 =AD=AC) and m(D)=30° and since m(BOC)=150° we get DBOC cyclic quadrilateral while A is the center of the circle. Hence AB=AO=AC radii of the same circle.

Migue:

AD perpendicular to BO

m(OCB) + m(OBC) = 30°

m(ODA) = 60°

Quadrilateral DBAC is concyclic

m(BAD) = m(OCB)

m(DBC) = m(DAC) = 60° - m(BAD) = 60° - m(OCB)

m(DBO) = m(DBC) - m(OBC) = 30°

Triangles DBO and ABO are isosceles. ==> AB = AO.

Fox 269

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Fox 268

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Fox 265 - Solutions

Joe has already posted a straightforward solution for Fox 265. Below are several more:

Pure Geometry - Little Trigonometry by Bleaug:
a=angle(CMD)=angle(CBD)=angle(AOC)
PP'=AA'=2.A'P' therefore tan(angle(PAB))=1/4. Take B' such that PB//P'B' then angle(A'P'B')=angle(PAB). By construction OA'=OB', so tan(a)=A'P'/OA'=2.A'P'/A'B'=2/tan(angle(A'P'B'))=8.

More Trigonometry by Yu:
Radius of the small circle is r, radius of the large circle is R.
tanθ = 1/4, tanΦ=tan2θ=2tanθ/(1-tanθ^2) = 8/15
so, r / sqrt(R^2 - r^2) = 8/15 => R = (17/8)r
tan* = r / (R - 2r) = 8.

Nice use of Pythagoras By Bob Ryden:Inscribed m(ABC) = (1/2) m(AOC) = m(POQ), so we can find m(POQ).
Let M be the midpoint of LN. Draw OM and ON.
Let radius of small circle = a.
Let PO = x.
Then radius of large circle = 2a + x.
OM = 2a – x.
MN = a.
ON = radius of large circle = 2a + x.
OM^2 + MN^2 = ON^2.
(2a – x)^2 + a^2 = (2a + x)^2
a = 8x
tan (ABC) = tan (POQ) = a/x = 8.

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Fox 267

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Note that the above is not the only configuration. This one can be classified as "intersection point of the lines can NEVER be in the interior of the equilateral triangle."

Thank you Lou for this nice problem.

Fox 266

This simple start will lead to a good one submitted by Lou.

There should be quite a few pure geometric solutions.

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Fox 261 - Solutions

Four good solutions for Fox 261, all drawing lines in the sand...

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Fox 261

The claim below makes some information in Fox 252 redundant.

http://www.8foxes.com/

This is an intriguing construction! We can easily post half a dozen more good foxes. Keep ckeckin' !!
Also there are at least 3 purely-geometric solutions...

Fox 256

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Hey you, tired of theory?

Start flying a kite.

Put aside the numbers,

Play like a child!

Fox 236 - Solution-1

See the original problem: Fox 236

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Let simplicity ring all accross the land...

Fox 243 - Solution

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Bleaug claims the following:

Almost a pure geometric proof this time (only requires Pythagorean theorem).

- TIJ is the section of the pyramid by a plane orthogonal to base and parallel to BC and AD

- by construction, TI is orthogonal to AB, and TJ to CD

- hence: AT^2-BT^2 = AI^2+IT^2-BI^2-IT^2 = AI^2-BI^2, similarly DT^2-CT^2 = DJ^2-CJ^2

- since AIB is a translation of DJC: AT^2-BT^2 = DT^2-CT^2

- from which we get DT^2 = AT^2-BT^2+CT^2 = 9-2+1 = 8

Fox 242

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A nice counterexample was found!

Fox 221 - Solution - 2

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