154700

1 Prove H=(G^2)/A
2 LHS:2/(1/b+1/a)
3 RHS:((sqrt(ab))^2)/((a+b)/2)
4 Simplifying RHS, we get
5 ab*(2/(a+b))
6 2ab/(a+b)
7 Dividing numerator and denominator by ab, we get
8 RHS=2/(1/b+1/a)=LHS
9 For part 2, it is clear that the length A is the radius of the semicircle
10 r=d/2 , therefore
11 A=(a+b)/2
12 Call the point where the side labelled A intersects the arc of the semicircle "P"
13 Call the point where the side labelled A intersects the diameter of the
14 semicircle "S".
15 Call the point where the side labelled G intersects the arc of the semicircle "Q"
16 Call the point where the side labelled G intersects the diameter of the
17 semicircle "R"
18 Call the intersection point of PR and QS "O"
19 Call the point where the perpendicular from R to OS intersects OS "X"
20 Triangles OPS and ORQ are similar, because the two angles at O are on
21 opposite sides of intersecting lines, meaning they are equal.
22 Observing that QS is a transversal crossing RS and QZ, such that QZ is
23 parallel to RS, gives an easy proof of the similarity of the two triangles
24 Call angle XQR alpha and it can be seen that angle XRQ is 90-alpha
25 Assuming angle QRS is a right angle, angle QSR is 90-alpha
26 Therefore, triangles QXR and QRS are similar. Thus we have the following
27 ratios
28 QS/QR=QR/QX
29 (QR)^2=QS*QX
30 QR=G, QS=A (it is a radius) and QX=H
31 Therefore, we have G^2=AH, which we proved earlier
32 Thus we have proved that A,G and H are equal to the three means
33 Consider the right triangle PSR. PR=Q. By the pythagorean theorem, we have
34 (SR)^2=(PR)^2-(SP)^2
35 (SR)^2=((a^2+b^2)/2)-((a^2+2ab+b^2)/4)
36 (SR)^2=(a^2-2ab+b^2)/4
37 SR=(a-b)/2
38 From the pythagorean theorem, in triangle QRS,
39 (SR)^2+(QR)^2=(QS)^2
40 LHS: ((a^2-2ab+b^2)/4)+ab=(a^2+2ab+b^2)/4
41 RHS: ((a+b)/2)^2 =(a^2+2ab+b^2)/4
42 LHS=RHS, so Q=sqrt((a^2+b^2)/2)