Fox 270

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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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