Given unit hexagon ABCDEF. Diagonals AC and BF intersect at G.
P is a point on CD at a distance of p from C.
Ray PG intersects AB at Q.
Find the ratio of the areas of triangles CGP and AGQ in terms of p.
Note for example that angle GBC is a right angle so that ABG is 30°. Angle GCP is also a right angle.
for p=.1:.1:.9
x=(sind(atand( p*sqrt(3)/2))/sqrt(3))/sind(150-atand(p*sqrt(3)/2));
A=sind(30)*x/(2*sqrt(3));
A1=p/sqrt(3);
disp([p A1/A])
end
In the above x is the measure of AQ, found using the law of sines, as being opposite an angle whose tangent is p*sqrt(3)/2 by being the opposite angle of angle PGC in that same named right triangle. The sine law is in conjunction with side AG being 1/sqrt(3) and the angle opposite AG in the small triangle being 180° - atan(p*sqrt(3)/2) - 30°.
The area of the small triangle, AQG, is sin(30°)*x*(1/sqrt(3)) / 2.
The area of the large triangle, GCP, is p/sqrt(3).
The table produced is:
p Area ratio
0.1 4.6
0.2 5.2
0.3 5.8
0.4 6.4
0.5 7
0.6 7.6
0.7 8.2
0.8 8.8
0.9 9.4
The measures on the GSP diagram are relative so the scale doesn't matter, but the absolute measurements farther to the left are then calculated relative to one of the sides.
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Posted by Charlie
on 2022-06-15 10:09:35 |
computer-aided solution (spoiler)