原問題[url]https://math.pro/db/viewthread.php?tid=1268&page=1#pid9288[/url]
[img]https://math.pro/db/attachment.php?aid=1970&k=995592d8c4327be6f1a1dc67316a8524&t=1380601096&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=1971&k=6eac1ff49bbda4d38ec029e888a72adb&t=1380601096]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=1268&page=1#pid7304[/url]
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[url=https://math.pro/db/attachment.php?aid=2007&k=def7c5dc9abc0d3a5f4211392fef1e69&t=1386862021]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=1853&page=2#pid9912[/url]
[img]https://math.pro/db/attachment.php?aid=2114&k=37eaea9a27b6622f5e8a3553218af59e&t=1397997236&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=2115&k=c97a11dd78364ebd375d0f0aeebcf303&t=1397997236]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=1868&page=1#pid10019[/url]
[img]https://math.pro/db/attachment.php?aid=2146&k=d45aa38eb3c67937fc42122e0154df92&t=1398517111&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=2145&k=bac3cd913aa60288853eac5004f0e646&t=1398517111]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=1872&page=1#pid10131[/url]
[img]https://math.pro/db/attachment.php?aid=2174&k=268c460e712f9c4c8519faebd115d9f3&t=1398817659&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=2175&k=2c34fd572840706cb6bb47c68889b723&t=1398817659]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=960&page=1#pid2178[/url]
[img]https://math.pro/db/attachment.php?aid=2543&k=d9e28a6d48e7d73a4f3c33addada100f&t=1410524096&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=2544&k=bea352b70e9e65568b79ad119615a2aa&t=1410524096]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=1268&page=1#pid13303[/url]
[img]https://math.pro/db/attachment.php?aid=2844&k=9b1045fc7a02d9938450f4c58fe9aef2&t=1431413750&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=2845&k=8d1d812d036e4906644c7b6094da6d92&t=1431413750]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=2154&page=3#pid14888[/url]
[img]https://math.pro/db/attachment.php?aid=3222&k=6c274797b2027d31b60e31ef3073fe31&t=1457211299&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=3223&k=e86f284d234b7fbb32d877c2a9272a67&t=1457211299]SketchUp檔下載[/url]
105.6.10補充
完成一件SketchUp作品需要以下幾個步驟
1.選題
評估該數學題目是否適合用SketchUp呈現，該採用靜態圖還是動態圖呢？
2.手畫分鏡圖
[img]https://math.pro/db/attachment.php?aid=3471&k=47e6f1e0e8e8cceedd7e6d1e90caa2b2&t=1465542506&noupdate=yes[/img]
3.SketchUp畫圖
[img=315,289]https://math.pro/db/attachment.php?aid=3472&k=b6aec815ac3b77db95326bdbd0c8abfa&t=1465542506&noupdate=yes[/img]
4.輸出bmp圖
[img=472,350]https://math.pro/db/attachment.php?aid=3473&k=fa472595da03a9ddad3f0488a5280db0&t=1465542506&noupdate=yes[/img]
5.bmp圖檔加上文字
[img=382,378]https://math.pro/db/attachment.php?aid=3474&k=703313f32a18591de76cb7efbfdc2209&t=1465542506&noupdate=yes[/img]
文字A,B移動時的座標
[img=238,402]https://math.pro/db/attachment.php?aid=3475&k=c6e701c4fb22de593f84bc90fd284cb6&t=1465542506&noupdate=yes[/img]
6.製成GIF
[img=568,468]https://math.pro/db/attachment.php?aid=3476&k=1d550fc0f3e5a04c150133a1b5319cbf&t=1465542506&noupdate=yes[/img]
原問題[url]https://math.pro/db/viewthread.php?tid=924&page=1#pid1968[/url]
[img=315,289]https://math.pro/db/attachment.php?aid=3722&k=87ceb15edbfa9d75bf6a6b7af74be643&t=1482538242&noupdate=yes[/img]
▲球面三角垛
[url=https://math.pro/db/attachment.php?aid=890&k=3d3d2c664d7e6dbf153fee7cb9b501d4&t=1326618156]SketchUp檔下載[/url]
[img=259,182]https://math.pro/db/attachment.php?aid=3723&k=eb6293ea1a796289af3b01e27d3a9361&t=1482538242&noupdate=yes[/img]
▲功能表 視窗/元件選項可以調整每個邊球的個數
[img=225,451]https://math.pro/db/attachment.php?aid=3724&k=f84eeddf69012c6f2a6c57de2384803d&t=1482538242&noupdate=yes[/img]
▲在SketchUp Pro版才能在功能表 視窗/元件屬性看到運作的機制
各屬性的公式如下
Copies=三角垛!Size*(三角垛!Size+1)*(三角垛!Size+2)/6-1
Discriminant=sqrt(9*(copy+1)*(copy+1)-1/27)
PositionZ=floor(-1+power(3*(copy+1)+Discriminant,0.33333)+power(3*(copy+1)-Discriminant,0.33333))
Remainder=copy-PositionZ*(PositionZ+1)*(PositionZ+2)/6
PositionX=floor((-1+sqrt(8*Remainder+1))/2)
PositionY=Remainder-PositionX*(PositionX+1)/2
X=(sqrt(3)*PositionX-2/sqrt(3)*PositionZ-1)*100
Y=(2*PositionY-PositionX-1)*100
Z=(2/3*sqrt(6)*(-1+三角垛!Size-PositionZ))*100
簡述背後的數學原理
1.要複製幾個球
Copies=三角垛!Size*(三角垛!Size+1)*(三角垛!Size+2)/6-1
每個球編號以SketchUp內建變數copy表示，編號從0到Copies。
三角垛個數公式為\( \displaystyle 1+3+6+10+\ldots+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6} \)
[url]https://zh.wikipedia.org/wiki/垛积术#.E4.B8.89.E8.A7.92.E5.9E.9B[/url]
2.編號第copy號的球在第幾層
Discriminant=sqrt(9*(copy+1)*(copy+1)-1/27)
PositionZ=floor(-1+power(3*(copy+1)+Discriminant,0.33333)+power(3*(copy+1)-Discriminant,0.33333))
從三角垛最上面的球編號下來
21
11 22 23
５ 12 13 24 25 26
２ ６ ７ 14 15 16 27 28 29 30
１ ３ ４ ８ ９ 10 17 18 19 20 31 32 33 34 35
第1層 第2層 第3層 第4層 第5層
編號從1到35個球，但SketchUp內建變數copy從0開始數，所以copy要+1才是對應的編號。
一元三次方程式\( \displaystyle \frac{Z(Z+1)(Z+2)}{6}=copy+1 \)
化簡後得到\( Z^3+3Z+2Z-6(copy+1)=0 \)
利用一元三次方程式的公式解\( ax^3+bx^2+cx+d=0 \)
判別式\( D=\sqrt{\displaystyle \left( \frac{bc}{6a^2}-\frac{b^3}{27a^3}-\frac{d}{2a} \right)^2+\left( \frac{c}{3a}-\frac{b^2}{9a^2} \right)^3} \)
\( \displaystyle x=-\frac{b}{3a}+\root 3 \of{\displaystyle \frac{bc}{6a^2}-\frac{b^3}{27a^3}-\frac{d}{2a}+D}+\root 3 \of{\displaystyle \frac{bc}{6a^2}-\frac{b^3}{27a^3}-\frac{d}{2a}-D} \)
[url]https://zh.wikipedia.org/wiki/三次方程#.E4.B8.89.E6.AC.A1.E6.96.B9.E7.A8.8B.E8.A7.A3.E6.B3.95[/url]
各項係數\( a=1 \)，\( b=3 \)，\( c=2 \)，\( d=-6(copy+1) \)
判別式\( D=\sqrt{\displaystyle 9(copy+1)^2-\frac{1}{27} } \)
\( Z=-1+\root 3 \of {3(copy+1)+D}+\root 3 \of {3(copy+1)-D} \)
\( PositionZ=\left[ Z \right] \)再取高斯函數得到整數
要注意的是PositionZ還要+1才是所在的層數
例如：copy=15，編號第16號球，計算得到PositionZ=3，但球在第4層
3.餘數多少
Remainder=copy-PositionZ*(PositionZ+1)*(PositionZ+2)/6
例如：copy=15，編號第16號球，PositionZ=3，計算得到Remainder=5(從0開始數)，在第6個位置
4.在第幾排的第幾個
PositionX=floor((-1+sqrt(8*Remainder+1))/2)
PositionY=Remainder-PositionX*(PositionX+1)/2
一元二次方程式\( \displaystyle \frac{Z(Z+1)}{2}=Remainder \)
化簡後得到\( Z^2+Z-2Remainder=0 \)
\( \displaystyle Z=\frac{-1+\sqrt{8Remainder+1}}{2} \)
\( PositionX=\left[ Z \right] \)再取高斯函數得到整數
例如：copy=15，編號第16號球，Remainder=5，計算得到PositionX=2(從0開始數)，在第3排
\( \displaystyle PositionY=Remainder-\frac{PositionX(PositionX+1)}{2}=2 \)(從0開始數)，左邊數過來第3個
5.轉換成球心坐標
X=(sqrt(3)*PositionX-2/sqrt(3)*PositionZ-1)*100
Y=(2*PositionY-PositionX-1)*100
Z=(2/3*sqrt(6)*(-1+三角垛!Size-PositionZ))*100
參考\(n=4\)各層球心坐標，就可以知道球心坐標公式
[url]https://math.pro/db/attachment.php?aid=3726&k=88b47f89cc9109aeecfdaee82ddc7e97&t=1482545551[/url]
補充資料
SketchUp動態元件教學
[url]http://help.sketchup.com/en/article/3000118[/url]
原問題[url]http://math.pro/db/viewthread.php?tid=924&page=3#pid2969[/url]
其中物件移動用到plug-in，但Keyframe Animation 2.0版只適用SketchUp 2014之後的版本，而我仍在用SketchUp 8，故使用Keyframe Animation 1.3完成作品。
Keyframe Animation for SketchUp
[url]http://regularpolygon.org/keyframe-animation/[/url]
[img]https://math.pro/db/attachment.php?aid=3804&k=a164f72b1e33029c2e7776406185e594&t=1488067850&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=3805&k=3607cfbaca176be958e2a2a9dc008b2d&t=1488067850]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=741&page=1#pid1294[/url]
[img]https://math.pro/db/attachment.php?aid=4246&k=4db1952d2cb874c1aa3e3e0a8c768c34&t=1502359021&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=4247&k=ea4b960636104e1a3481f13c3e3ec893&t=1502359021]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/thread-2927-1-1.html[/url]
[img]https://math.pro/db/attachment.php?aid=4345&k=c8b59e97c0ce43f18259a4251ad5c030&t=1522971315&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=4335&k=9674d7a89747170688d8f035a5d755a5&t=1529246330]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=2943&page=2#pid18354[/url]
[img]https://math.pro/db/attachment.php?aid=4591&k=058aff3ec572d093fb49c5a215af0a42&t=1529245927&noupdate=yes[/img]
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原問題[url]https://math.pro/db/viewthread.php?tid=2943&page=2#pid18355[/url]
[img]https://math.pro/db/attachment.php?aid=4589&k=ae17ccebbf185ecea384f05f67a09d6d&t=1529245927&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=4590&k=94c68da7452ff79761804fc7f7de082b&t=1529245927]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=2927&page=1#pid19058[/url]
[img]https://math.pro/db/attachment.php?aid=4704&k=f20ff67f6153ee3be503b8e59842858d&t=1537693174&noupdate=yes[/img]
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[img]https://math.pro/db/attachment.php?aid=4706&k=b59b85fb40f12323ae0288dff2055c24&t=1537693174&noupdate=yes[/img]
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原問題[url=https://math.pro/db/viewthread.php?tid=3132&page=1#pid19867]https://math.pro/db/viewthread.php?tid=3132&page=1#pid19867[/url]
[img]https://math.pro/db/attachment.php?aid=5134&k=85dea65063f342890901584f6a8ea46f&t=1559997142&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=5135&k=fd7cc9e2b20366496b756fcfeca290d1&t=1559997142]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=1003&page=2#pid14609[/url]
邊長13,14,15的三角形相關計算如下
設\(\Delta ABC\)各頂點坐標為\(\displaystyle A(0,0),B(15,0),C(\frac{33}{5},\frac{56}{5})\)
\(\Delta ABC\)的內切圓圓心\(I(7,4)\)，切\(\overline{AB}\)於\(F(7,0)\)，切\(\overline{BC}\)於\(\displaystyle D(\frac{51}{5},\frac{32}{5})\)，切\(\overline{CA}\)於\(\displaystyle E(\frac{231}{65},\frac{392}{65})\)
因為SketchUp的圓只是用正24邊形代替，當你在SketchUp操作以\(A\)點為圓心，半徑13畫圓和以\(B\)點為圓心，半徑14畫圓，設兩圓的交點為\(C\)，但測量線段長度得\(\overline{AC}=12.97,\overline{BC}=13.39\)，\(\Delta ABC\)形狀已經有誤差，直接計算\(C\)點坐標。
解聯立方程式\(\cases{x^2+y^2=13^2 \cr (x-15)^2+y^2=14^2}\)，得到\(\displaystyle C(\frac{33}{5},\frac{56}{5})\)
SketchUp也沒有畫角平分線工具，直接計算\(\Delta ABC\)內心坐標。
\(\displaystyle \vec{OI}=\frac{a}{a+b+c}\vec{OA}+\frac{b}{a+b+c}\vec{OB}+\frac{c}{a+b+c}\vec{OC}\)
\(\displaystyle =\frac{14}{13+14+15}(0,0)+\frac{13}{13+14+15}(15,0)+\frac{15}{13+14+15}(\frac{33}{5},\frac{56}{5})\)
\(=(7,4)\)
得到內心坐標\(I(7,4)\)
以SketchUp所畫出的內切圓和三角形三邊的交點也會有誤差，直接計算交點坐標
\(\overline{BC}\)直線方程式為\(\displaystyle y-0=\frac{0-11.2}{15-6.6}(x-15)\)，\(4x+3y=60\)
\(\overline{ID}\)直線方程式為\(\displaystyle y-4=\frac{3}{4}(x-7)\)，\(3x-4y=5\)
解聯立方程式\(\cases{4x+3y=60 \cr 3x-4y=5}\)，交點\(\displaystyle D(\frac{51}{5},\frac{32}{5})\)
\(\overline{AC}\)直線方程式為\(\displaystyle y-0=\frac{11.2-0}{6.6-0}(x-0)\)，\(56x-33y=0\)
\(\overline{IE}\)直線方程式為\(\displaystyle y-4=-\frac{33}{56}(x-7)\)，\(33x+56y=455\)
解聯立方程式\(\cases{56x-33y=0 \cr 33x+56y=455}\)，交點\(\displaystyle E(\frac{231}{65},\frac{392}{65})\)
內切圓和\(\overline{AB}\)交點為\(F(7,0)\)
[img]https://math.pro/db/attachment.php?aid=5757&k=4d2a2b049c34cdc12078f9e35e056e22&t=1613484826&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=5758&k=ab84c86a02072a7df4b1fb54c7e6badf&t=1613484826]SketchUp檔下載[/url]
[img]https://math.pro/db/attachment.php?aid=5760&k=ff348b4f46dfa5d19c354276702dbde7&t=1613484826&noupdate=yes[/img]
[url=https://math.pro/db/attachment.php?aid=5761&k=df5c8b13a1acf77471412b4d1284d8d4&t=1613484826]SketchUp檔下載[/url]
原問題[url]https://math.pro/db/viewthread.php?tid=3490&page=1#pid22082[/url]
使用maxima計算結果如下
設半徑4的球心\(A(0,0,0)\)，半徑9的球心\(B(13,0,0)\)，半徑16的球心\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)
外公切平面方程式為\(30\sqrt{29}x+271y\pm 13\sqrt{455}z+312\sqrt{29}=0\)
和半徑4的球相切於\(\displaystyle A' \left(-\frac{20}{13},-\frac{542}{39\sqrt{29}},-\frac{2\sqrt{455}}{3\sqrt{29}}\right)\)
和半徑9的球相切於\(\displaystyle B' \left(\frac{124}{13},-\frac{813}{26\sqrt{29}},-\frac{3\sqrt{455}}{2\sqrt{29}}\right)\)
和半徑16的球相切於\(\displaystyle C' \left(-\frac{108}{13},\frac{2008}{39\sqrt{29}},-\frac{8\sqrt{455}}{3\sqrt{29}}\right)\)
有前一次的經驗後就直接計算各點位置
設半徑4的球心\(A(0,0,0)\)，半徑9的球心\(B(13,0,0)\)，半徑16的球心\(C(x,y,0)\)
\(\cases{\overline{BC}=25\cr \overline{CA}=20}\)，\(\cases{(x-13)^2+(y-0)^2=25^2 \cr (x-0)^2+(y-0)^2=20^2}\)，
\(\displaystyle x=-\frac{28}{13},y=\frac{48\sqrt{29}}{13}\)，得到\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)
設\(P(x,y,0)\)為兩個外公切平面方程式交線上一點
球心\(A(0,0,0)\)半徑4的球面和平面相切於\(A'\)點，\(\Rightarrow \overline{AA'}=4\)，\(∠PA'A=90^{\circ}\)
球心\(B(13,0,0)\)半徑9的球面和平面相切於\(B'\)點，\(\Rightarrow \overline{BB'}=9\)，\(∠PB'B=90^{\circ}\)
\(\Delta PAA'\)和\(\Delta PBB'\)為相似三角形(\(∠APA'=∠BPB'\)，\(∠PA'A=∠PB'B=90^{\circ}\))
\(\overline{PA}:\overline{PB}=\overline{AA'}:\overline{BB'}=4:9\)，\(\overline{PA}:\overline{AB}=4:5\)
由外分點公式可知\(\displaystyle P=\frac{9}{5}A-\frac{4}{5}B=\frac{9}{5}(0,0,0)-\frac{4}{5}(13,0,0)=(-\frac{52}{5},0,0)\)
得到\(\displaystyle P(-\frac{52}{5},0,0)\)
設\(Q(x,y,0)\)為兩個外公切平面方程式交線上一點
球心\(A(0,0,0)\)半徑4的球面和平面相切於\(A'\)點，\(\Rightarrow \overline{AA'}=4\)，\(∠PA'A=90^{\circ}\)
球心\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)半徑16的球面和平面相切於\(C'\)點，\(\Rightarrow \overline{CC'}=16\)，\(∠PC'C=90^{\circ}\)
\(\Delta PAA'\)和\(\Delta PCC'\)為相似三角形(\(∠APA'=∠CPC'\)，\(∠PA'A=∠PC'C=90^{\circ}\))
\(\overline{QA}:\overline{QC}=\overline{AA'}:\overline{CC'}=4:16\)，\(\overline{QA}:\overline{AC}=1:3\)
由外分點公式可知\(\displaystyle Q=\frac{4}{3}A-\frac{1}{3}C=\frac{4}{3}(0,0,0)-\frac{1}{3}C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)=\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)
得到\(\displaystyle Q\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)
從\(\displaystyle P\left(-\frac{52}{5},0,0\right)\)和\(\displaystyle Q\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)求交線的對稱比例式
方向向量為\(\displaystyle \vec{PQ}=\left(\frac{2168}{195},-\frac{16\sqrt{29}}{13},0\right)\)
\(PQ\)直線的對稱比例式為\(\displaystyle \frac{x+\frac{52}{5}}{\frac{2168}{195}}=\frac{y-0}{-\frac{16\sqrt{29}}{13}}\)，\(z=0\)
化簡得到\(30\sqrt{29}x+271y+312\sqrt{29}=0\)，\(z=0\)
假設外公切平面方程式為\(30\sqrt{29}x+271y+312\sqrt{29}+kz=0\)，\(k\in R\)
球心\(A(0,0,0)\)到平面距離\(\displaystyle \frac{|\;0+0+0+312\sqrt{29}|\;}{\sqrt{(30\sqrt{29})^2+271^2+k^2}}=4\)，\(k=\pm 13\sqrt{455}\)
得到外公切平面方程式\(30\sqrt{29}x+271y\pm 13\sqrt{455}z+312\sqrt{29}=0\)
利用投影點公式求切點\(A',B',C'\)
投影點公式：
\(P(x_0,y_0,z_0)\)對平面\(ax+by+cz+d=0\)的投影點為\(P'(x_0-at,y_0-bt,z_0-ct)\)，其中\(\displaystyle t=\frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}\)
計算\(A(0,0,0)\)的投影點\(A'\)
\(\displaystyle t=\frac{0+0+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{2}{39\sqrt{29}}\)
\(\displaystyle A'\left(0-30\sqrt{29}\cdot \frac{2}{39\sqrt{29}},0-271\cdot \frac{2}{39\sqrt{29}},0-13\sqrt{455}\cdot \frac{2}{39\sqrt{29}}\right)=
\left(-\frac{30}{13},-\frac{542}{39\sqrt{29}},-\frac{2\sqrt{455}}{3\sqrt{29}}\right)\)
計算\(B(13,0,0)\)的投影點\(B'\)
\(\displaystyle t=\frac{30\sqrt{29}\cdot 13+0+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{3}{26\sqrt{29}}\)
\(\displaystyle B'\left(13-30\sqrt{29}\cdot \frac{3}{26\sqrt{29}},0-271\cdot \frac{3}{26\sqrt{29}},0-13\sqrt{455}\cdot \frac{3}{26\sqrt{29}}\right)=
\left(\frac{124}{13},-\frac{813}{26\sqrt{29}},-\frac{3\sqrt{455}}{2\sqrt{29}}\right)\)
計算\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)的投影點\(C'\)
\(\displaystyle t=\frac{30\sqrt{29}\cdot (-\frac{28}{13})+271\cdot \frac{48\sqrt{29}}{13}+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{8}{39\sqrt{29}}\)
\(\displaystyle C'\left(-\frac{28}{13}-30\sqrt{29}\cdot \frac{8}{39\sqrt{29}},\frac{48\sqrt{29}}{13}-271\cdot \frac{8}{39\sqrt{29}},0-13\sqrt{455}\cdot \frac{8}{39\sqrt{29}}\right)=
\left(-\frac{108}{13},\frac{2008}{39\sqrt{29}},-\frac{8\sqrt{455}}{3\sqrt{29}}\right)\)
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