Fox 287

We may have a single solution, but not sure if that's the whole story.

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Fox 285 - Solution

Bleaug constructs:
Here is a geometric construction (i.e. compass and ruler) of an inscribed triangle with required property for any angle from 0 to 40°:
1- take any point b on AB and build point c such as angle(Abc)=2α
2- build a inside ABC such as abc is equilateral
3- build O such that angle(Oca)=angle(Oba)=α
4- build P intersection of AO and BC
5- build PQR homothetic to Obc

PQR is such that angle(BPQ)=α, angle (AQR)=2α, angle(CRP)=3α

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Fox 285

GEOMETRIC CONSTRUCTION - A Beautiful Greek Antiquity:
(From Weisstein, Eric W. "Geometric Construction." From MathWorld--A Wolfram Web Resource.)
In antiquity, geometric constructions of figures and lengths were restricted to the use of only a straightedge and compass (or in Plato's case, a compass only; a technique now called a Mascheroni construction). Although the term "ruler" is sometimes used instead of "straightedge," the Greek prescription prohibited markings that could be used to make measurements. Furthermore, the "compass" could not even be used to mark off distances by setting it and then "walking" it along, so the compass had to be considered to automatically collapse when not in the process of drawing a circle.

Because of the prominent place Greek geometric constructions held in Euclid's Elements, these constructions are sometimes also known as Euclidean constructions. Such constructions lay at the heart of the geometric problems of antiquity of circle squaring, cube duplication, and angle trisection. The Greeks were unable to solve these problems, but it was not until hundreds of years later that the problems were proved to be actually impossible under the limitations imposed. (Also see Compass and straightedge constructions)

This is our first geometric construction problem, based on a solution submitted by Bleaug. We have a few other construction problems in the queue. For those who are new: the question asks to draw triangle PQR in any equilateral triangle by using a compass, a ruler (without any numbers on it), and angle α (and a piece of paper too :) Let us know if you have any questions!

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Fox 284

Another simple one in the series.

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Fox 276 - Solution

Solution to 276 was already known but figure below explains more. By Jim Wilson:

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The locus of point P is the arc of a circle. Because the central angle must be twice the subtended arc in measure, the central angle is 60. Thus the major arc has length 10π/6. Since the trajectory would also be over a second arc that is a reflection in AB, the total trajectory is 10π/3.

Fox 276

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Fox 196 - Solution 1

We have received another great solution for Fox 196, submitted by Janos Pataki.