Euclidea 14.5 解法与证明

Euclidea 14.5

如图，给定三个两两相切的圆，O，O_1， O_2。三个圆心共线。

任务：尺规作出一个圆，与给定的三个圆都相切。

解法：

设圆O与圆O_1的切点为A
连接OA
过O做OA垂线交圆O于D
以D为圆心，DA为半径作圆，交圆O_1和圆O_2于E和F。
作直线EO_1和FO_2交于G
以G为圆心，GE为半径作圆
圆G即为所求圆

简要证明：

如图，设三个圆切点为A，B，C。明显A，B，C，O，O_1，O_2五点共线。
以A为中心作反演变换，取反演圆与圆O_2正交。于是圆O_2经反演变换后不变。B，C互为反演点。
分别过B和C做AB的垂线L_B和L_C。易知圆O1和圆O经反演变换后分别为L_B和L_C。
如图作圆G'同时与圆O_2，L_B和L_C相切于E'和F'。
过E'F'作直线L，易证C在L上，并且∠E'CB为45°。
设L经过反演变换得到的圆为圆D'，考虑圆D'的性质：

直线AD'与直线L垂直
圆D'经过A（因为L不经过A）
圆D'经过B（因为L经过C，而B，C互为反演点）

于是可知圆D就是圆D'，因此E和E', F和F'分别互为反演点。
设圆G'经过反演变换为圆G''。因为圆G'与L_B和圆O_2相切于E'和F'，所以圆G''与圆O_1和圆O_2相切于E和F。
所以G''即是EO_1和FO_2的交点，且圆G''半径是G''E。于是圆G即是圆G''。
因为圆G'与L_B，圆O_2，L_C都相切，所以圆G与三个给定圆都相切。证明完毕。

参考资料：Pappus chain

Comments

Determine Perspective Lines With Off-page Vanishing Point

In perspective drawing, a vanishing point represents a group of parallel lines, in other words, a direction. For any point on the paper, if we want a line towards the same direction (in the 3d space), we simply draw a line through it and the vanishing point. But sometimes the vanishing point is too far away, such that it is outside the paper/canvas. In this example, we have a point P and two perspective lines L1 and L2. The vanishing point VP is naturally the intersection of L1 and L2. The task is to draw a line through P and VP, without having VP on the paper. I am aware of a few traditional solutions: 1. Use extra pieces of paper such that we can extend L1 and L2 until we see VP. 2. Draw everything in a smaller scale, such that we can see both P and VP on the paper. Draw the line and scale everything back. 3. Draw a perspective grid using the Brewer Method. #1 and #2 might be quite practical. #3 may not guarantee a solution, unless we can measure distances/p...

Chasing an IO Phantom

My home server has been weird since months ago, it just becomes unresponsive occassionally. It is annoying but it happens only rarely, so normally I'd just wait or reboot it. But weeks ago I decided to get to the bottom of it. What's Wrong My system set up is: Root: SSD, LUKS + LVM + Ext4 Data: HDD, LUKS + ZFS 16GB RAM + 1GB swap Rootless dockerd The system may become unresponsive, when the IO on HDD is persistantly high for a while. Also: Often kswapd0 has high CPU High IO on root fs (SSD) From dockerd and some containers RAM usage is high, swap usage is low It is very strange that IO on HDD can affect SSD. Note that when this happens, even stopping the IO on HDD does not always help. Usually restarting dockerd does not help, but rebooting helps. Investigation: Swap An obvious potential root cause is the swap. High CPU on kswapd0 usually means the free memory is low and the kernel is busy exchanging data between disk and swap. However, I tried the following steps, none of the...

Moving Items Along Bezier Curves with CSS Animation (Part 2: Time Warp)

This is a follow-up of my earlier article. I realized that there is another way of achieving the same effect. This article has lots of nice examples and explanations, the basic idea is to make very simple @keyframe rules, usually just a linear movement, then use timing function to distort the time, such that the motion path becomes the desired curve. I'd like to call it the "time warp" hack. Demo See the Pen Interactive cubic Bezier curve + CSS animation by Lu Wang ( @coolwanglu ) on CodePen . How does it work? Recall that a cubic Bezier curve is defined by this formula : \[B(t) = (1-t)^3P_0+3(1-t)^2tP_1+3(1-t)t^2P_2+t^3P_3,\ 0 \le t \le 1.\] In the 2D case, \(B(t)\) has two coordinates, \(x(t)\) and \(y(t)\). Define \(x_i\) to the be x coordinate of \(P_i\), then we have: \[x(t) = (1-t)^3x_0+3(1-t)^2tx_1+3(1-t)t^2x_2+t^3x_3,\ 0 \le t \le 1.\] So, for our animated element, we want to make sure that the x coordiante (i.e. the "left" CSS property) is \(...

WangLu's Notes

Search This Blog