Key words: Geometry, Unusual geometry, Math, Physics, Chemistry, High school, Geometry Olympiads, Free Geometry, Euclidean Geometry, Calculus, Geometric Construction. Oh yes, going-nowhere discussions, haikus, and poems too.

## Wednesday, March 31, 2010

### Fox 270

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## Monday, March 29, 2010

### Fox 266 - Solutions

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

__Bleaug:__

**Given phi=30°, then m(BOC) = 150° which is 1/2 of reflex m(BAC).**

__Yu:__.:B, O and C are on the same circle centred at A. Hence AB = AO.

**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.**

__Giannno:__

__Migue:__Reactions: |

### Fox 269

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## Friday, March 26, 2010

### Fox 268

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## Wednesday, March 24, 2010

### Fox 265 - Solutions

__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.

**:Inscribed m(ABC) = (1/2) m(AOC) = m(POQ), so we can find m(POQ).**

__Nice use of Pythagoras By Bob Ryden__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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## Monday, March 22, 2010

### Fox 267

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."

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## Friday, March 19, 2010

### Fox 266

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### Fox 257 - Solution

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## Thursday, March 18, 2010

### Fox 260 - Solutions

**Almost Pure Geometric Solution (An Essential Property of a Parabola)**

**by Bleaug**

Let's take as a parabola "well known" property that for any two points U, V of abscissa u, v, the tangent to point T of abscissa (u+v)/2 is parallel to vector UV. (OK! this demonstration would need some cartesian algebra but simple, promise!) From this we derive that for any point M between U and V the area of triangle UMV et less or equal to area of triangle ATV which maximizes triangle height (e.g. assume the opposite and compare area of triangle obtained from M+dM)

Then let's assume U and V achieve the maximum trapezoid area between A (x=0) and B(x=2), then necessarily Au=uv and uv=vB which implies Au=uv=vB=AB/3=2/3. Because of symmetry, maximum area is equivalent to area of rectangle AvVW = 8/9 * 4/3 = 32/27.**Geometric Translationby Yu**

Translate y=2x-x^2 to the left by 1 unit to obtain y=1-x^2.

Without going into details, the area of the trapezium is greater than the area of the quadrilateral. Area of trapezium, A = (1/2) (2x+2)(1-x^2) = (x+1)(1-x^2)

Max A = 32/27 when x = 1/3.

For more details see Fox 260.

http://www.8foxes.com/

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## Wednesday, March 17, 2010

### Fox 265

Red fox: Little changes lead nowhere! What we need is a whole scale revolution.

Polar fox: What are you suggesting? Rewriting Euclid's axioms?

Red fox: Why not? Define a new basis, set up a new geometry and build a new universe.

Polar fox: Dude, you're beyond here. I think you need a change.

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## Tuesday, March 16, 2010

### Fox 264

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## Sunday, March 14, 2010

### Fox 71

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### Fox 261 - Solutions

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## Friday, March 12, 2010

### Fox 263

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## Wednesday, March 10, 2010

### Fox 262

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## Tuesday, March 9, 2010

### Fox 261

Also there are at least 3 purely-geometric solutions...

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## Monday, March 8, 2010

### Fox 260

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## Saturday, March 6, 2010

### Fox 12

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## Friday, March 5, 2010

### Fox 11

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### Fox 255 - Solutions

__Median Formula by Wilson:__

__Heron's Formula by Wilson:__Reactions: |

## Thursday, March 4, 2010

### Fox 259

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## Wednesday, March 3, 2010

### Fox 258

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## Tuesday, March 2, 2010

### Fox 257

http://www.8foxes.com/

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### Fox 256 Solution

= area of parallelogram ABEF = area of parallelogram ABCD

http://www.8foxes.com/

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## Monday, March 1, 2010

### Fox 256

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