First note that triangle ABD and triangle BEF are similar. Also note that Be = s
ID: 2833126 • Letter: F
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First note that triangle ABD and triangle BEF are similar. Also note that Be = s4/2 - s8/2. Use these facts to show that Next note that triangle ACF and triangle CFG are similar. Use this fact to show that s8/2/t8=t8/t4 or t82=s8t4/2 Having related the side lengths of the various polygons, we now relate the perimeters of the polygons. We let S4 = 4s4 be the perimeter of the circumscribed square, t4 = 4t4 be the perimeter of the inscribed square. S8= 8s8 be the perimeter of the circumscribed octagon, and T8 = 8t8 be the perimeter of the inscribed octagon. Use these definitions in Steps 3 and 4 to show that s8=2s4t4/s4+t4 and t8= t4s8 Find S8 and 7g. the hard work is finished. To double the number of sides again (from 8 to 16 or from 16 to 32), the geometry of Figure 2 still applies. So if we know sn and tn for any n, the perimeters of the inscribed and circumscribed 2n-gons arc S2n = 2sntn/sn+tn and t2n= tns2n Note that S& is needed to compute so must be computed first. Beginning with s4= 4, and T4= 2 2 from Step 1, use the relations in Step 7 to complete the following table. One way to obtain a better approximation is to average sn and Tn. Include these averages in the table as well. It appears that Archimedes started with hexagons (n= 6) and computed perimeters of inscribed and circumscribed 12-gons, 24-gons, 48-gons, and 96-gons, which produced the approximation 3 10/71Explanation / Answer
3. As ABD and BEF are similar, |AB|/|AD| = |EB|/|EF|
|AB| = 1/2 s4; |AD| = 1/2 t4; |EB| = s4/2 - s8/2; |EF| = 1/2 s8
Thus, (1/2s4)/(1/2 t4) = (s4/2 - s8/2)/(1/2s8)
Just multiply both sides by 2/2
s4/t4 = (s4 - s8)/s8
To solve for s8, cross-multiply:
s8 s4 = t4(s4-s8)
s8 s4 = t4 s4 - t4 s8
s8 s4 + t4 s8 = t4 s4
s8(s4+t4) = t4 s4
s8 = s4t4/(s4+t4)
4. ACF and CFG are similar; thus |FG|/|FC| = |AF|/|AC|
|FG| = 1/2 s8; |FC| = t8; |AF| = t8; |AC| = t4;
Thus, (1/2 s8)/t8 = t8/t4, or (s8/2) /t8 = t8/t4
Cross-multiplying t8^2 = s8t4/2
5. 8s8 = S8; 8t8=T8; 4s4 = S4; 4t4 = T4;
From equation 3,
s8 = s4t4/(s4+t4)
Multiply the left and right sides by 8
8s8 = 8s4 t4/(s4+t4)
S8 = 2*4s4 t4/(s4+t4)
S8 = 2 S4 t4/(s4+t4)
multiply the right side by 4/4
S8 = 2S4 *4t4/(4s4+4t4)
S8 = 2S4T4/(S4+T4)
From 4,
t8^2 = s8t4/2
Multiply both sides by 64
64t8^2 = 64s8t4/2
64t8^2 = 32s8t4
8*8*t8*t8 = 8*4*s8*t4
8t8*8t8 = 8s8*4t4
T8*T8 = S8*T4
T8^2 = S8T4
T8 = sqrt(S8T4)
6. As S4 = 4 and T4 = 2 sqrt(2)
S8 = 2S4T4/(S4+T4) = 2*4*2 sqrt(2)/(4 + 2 sqrt(2)) = 16 sqrt(2)/(4 + 2 sqrt(2)) =
(multiplying by (4 - 2 sqrt(2))/(4-2sqrt(2)))
(64 sqrt(2) - 64)/(16 - 8) = (64 sqrt(2) - 64)/8 = 8 sqrt(2) - 8 = 3.31370849898476
T8 = sqrt(T4S8) = sqrt(2 sqrt(2) (8 sqrt(2) - 8)) = sqrt(32 - 16 sqrt(2)) = 4 sqrt(2 - sqrt(2)) = 3.06146745892072
7 and 8 (see error at bottom of the table)
S256 has an error of 0.000157715579 and T256 has an error of -0.000078852445
Not asked for, the error for the average is 0.000039431567 The error of S256 is nearly exactly double that of T256.
(this suggests that if we had used a weighted average of S256 and T256, we would be amazingly close; if you use
(S256 + 2 T256)/3, your error is 3.56273055501788E-09; much more sophisticated work would be necessary to show why this weighted average is a good idea.
9. If you look at the change in the error as you increase n, you will notice that it tends to a ratio of 1/4 in each step. Then, as S has a greater error, if we look at 0.000157715579 and divide by 10^-10, we get 1577155.
log 4 1577155 = 10.2944468722911
Then, let's use 11 steps from 256, which takes us to 256 * 2^11 = 524288
It turns out that this was exactly the right number of steps. Of course, 10 additional steps would have been sufficient for Tn and the average.
The weighted average I proposed, (Sn + 2Tn)/3, gave an error too small for Excel to calculate.
n Sn Tn (Sn+Tn)/2 4 4.000000000000 2.828427124746 3.414213562373 8 3.313708498985 3.061467458921 3.187587978953 16 3.182597878075 3.121445152258 3.152021515166 32 3.151724907429 3.136548490546 3.144136698988 64 3.144118385246 3.140331156955 3.142224771100 128 3.142223629942 3.141277250933 3.141750440438 256 3.141750369169 3.141513801144 3.141632085157 Error 0.000157715579 -0.000078852445 0.000039431567Related Questions
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