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[生活] 老封:平面几何热线

这是上回那题的证明吧?.

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这种证明方法是我现在教的一个高中竞赛班的叫徐文浩的学生给出的。

你知道他花了多少时间想这个问题?1分钟!呵呵,像在做口答题。

这个小子初三得过全国初三竞赛一等奖,高二得了高三竞赛二等奖,确实蛮厉害的。

这次我们学校参加ti杯比赛,我在决定打团体赛的名单时在两个人之间犹豫不决,最终放弃了一个比他稳定的,选择了他,算是赌博吧。:).

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我认为这种方法应该算是最简单的一种了,你说呢?.

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好样的!.

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老姜,你昨天给我思考的这题晚上已找到了推广,改造成为一道证明题。
你看看:.

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你昨晚的电话搅了我的清梦。不过,接这个电话还是有所收获的。希望能把这个问题的几何背景搞清楚,三角的面孔应该只是它的显像部分。.

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这两天正忙于备课,没时间想下去了。.

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据说解题高手唐传发老师还未正式登陆过,只是浏览了一下而已。

[ 本帖最后由 老封 于 2007-5-31 13:59 编辑 ].

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引用:
原帖由 老姜 于 2007-5-10 10:07 发表
你昨晚的电话搅了我的清梦。不过,接这个电话还是有所收获的。希望能把这个问题的几何背景搞清楚,三角的面孔应该只是它的显像部分。
你注意到没有:这个图两个角地位还是不对称的。
要想几何上论证,首先得把结论改造得更对称些 !
能不能把这个三角关系转化掉?.

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刚收到老殿的邮件:
老封你好:我以前讨论过Malfatti问题,并给出了相应的解法,好像《小花》中就有。
近日阅读吴文俊老先生的一篇《数学机械化:回顾与展望》,参见:
http://www.mmrc.iss.ac.cn/973/wu-pref.htm
其中有一段使我百思不解:
几何作图也是一类诱人的问题,在19世纪中叶得到充分的关注。除了通常的规尺作图外,19世纪的几何学家,还阐发了只用直尺或只用圆规之类的作图理论与作图方法。在古希腊时代,就有求作一圆与三圆相切的Appolonius问题以及所谓几何三大问题。19世纪又出现了所谓求作三圆彼此相切且各与三角形的两边相切的Malfatti问题,更重要的是给出了可以规尺作图的充要条件。例如Appolonius问题可以用规尺作出,而Malfatti问题则否。Gauss更据以证明可以规尺作图的所有可能的正多边形,特别指出正 17边形可以用规尺作出,这一出人意表的成果使年轻的Gauss决定献身数学。在近代,也有源自著名数学家Zassenhaus与Van der Waerden的一个问题。已知一个三角形的三条边,就可作出它的内外分角线来。反过来,知道三角形的内外分角线的三条,是否可以作出相应的三角形来,就很不简单,但运用上述判准,却可以得到完全的解决,即一般说来光用规尺是不可能的。
难道Malfatti问题真是尺规作图不能问题,还是下面作法有问题:先定出三线形a1a2a3内心I,再定出三角形A1A2I、A2A3I、A3A1I的内心I1、I2、I3,以I2I3、I3I1、I1I2为对称轴,分别将A1I、A2I、A3I反射变换得到共点的三条直线b1'、b2'、b3',则四线形a1a2b1'b2'、a2a3b2'b3'、a3a1b3'b1'的内切圆即为所求圆。请你看看,若Malfatti问题真是已经证明了的尺规作图不能问题,以上作法又没有问题,那问题可就大了:代数证明的尺规作图不能问题是不可靠的!弟:殿林  07-05-08


老殿,我想吴老先生在这一细节上出了点小错。Malfatti问题并不是尺规不能问题,作法不仅《小花》中有,记得阿达玛《几何》一书附录中还有详细讨论。看来对名人的言论也不能迷信啊!

[ 本帖最后由 老封 于 2007-5-14 10:15 编辑 ].

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最新成果与大家分享

昨晚在研究一个颇有意思的图形:

任意直线L关于△ABC三边的轴对称直线L1、L2、L3,它们围成一个三角形,记为△XYZ。我发现了两个有趣的结论——

结论一  △XYZ的内心I恰位于△ABC的外接圆上。.

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结论二  △XYZ的外接圆直径D与△ABC的垂心H离开直线L的距离d之比,恰好等于△ABC与垂三角形△DEF的面积之比!.

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还有,当直线L平移或绕定点旋转时,可观察到△XYZ的外心O′的轨迹分别是直线和圆,有意思的是,它们都经过△ABC外接圆上一个固定的点,它并不受直线L的方向或旋转中心的具体位置干扰。
那么这个固定的点有什么特点呢?
现在已经搞清楚了,该点的特点是——它对于△ABC的西摩松线恰好平行于△ABC的欧拉线!
后又发觉,当直线L平移或绕定点旋转时,△XYZ的垂心或其它特殊点的轨迹也都是直线或圆,而且也相应地经过定点。对这类定点的刻划是一件有意思的事。

[ 本帖最后由 老封 于 2007-5-15 09:13 编辑 ].

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又如,当直线L绕某个定点旋转时,可观察到△XYZ的欧拉线也相应地在绕另一个定点旋转。能不能把这两个定点之间的联系刻划出来?.

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这个问题就更为复杂了。
三角形XYZ(我记为A'1A'2A'3)的许多特殊点,当直线L平移时共点,该点就是它的内心I’(不动点)。在L绕定点旋转时,其欧拉线经过的定点是外心O‘、垂心H’轨迹圆的一个交点Q,另一个交点P在三角形A1A2A3的外接圆上(即三圆共点,也是XYZ的许多特殊点的轨迹——圆——的公共点)


[ 本帖最后由 老殿 于 2007-5-16 08:46 编辑 ].

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引用:
原帖由 老殿 于 2007-5-16 08:40 发表
这个问题就更为复杂了。
三角形XYZ ...
我的感觉与你不同,这个构形比作轴对称点的反而简单,而且意思更大。
从表面看,这题涉及的是作轴对称线,其实不然。从更深处看,它更多体现的是三相似图形的思想(而且还是一种特例)。其实一切都是在顺相似的范畴内做文章,所以结论更显和谐。下次有空我再对此作些解释。.

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我昨天提到了外心、垂心等特殊点的轨迹,其实特殊点并不是问题的关键。只要一个点在三条对称线所围成的三角形(我也改称其为△D′E′F′)中的相对位置保持不变,那么其轨迹就是一个圆,而且不管平移方向为何,旋转中心为何,它总与△ABC的外接圆交于一个定点,这个定点的位置是由所取点在△D′E′F′中的相对位置决定的。对此,我已给出了一个完美的结论,使问题得以最终解决。
不过这个结论在外人看来或许有些费解,未必能看出其深刻之处。但想必也一定有人能感受到它的价值:.

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结论三    设P′是△ABC外接圆上任一点,过垂心H作P点的西摩松线的平行线,并在这条平行线上取两点P和Q(在H点的两侧),满足:│PH│×│HQ│=2Rr,其中,R是△ABC外接圆半径,r是△DEF内切圆的半径。再过Q作任意直线L,记L关于△ABC三边的轴对称直线所围成三角形为△D′E′F′。那么,结论是:D′、E′、F′、P′四点一定与D、E、F、P四点反向相似!.

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补注:在结论三中,2Rr也等于AH×HD(或BH×HE及CH×HF)。

有了结论三,就把轨迹圆所经过的那个神秘的不动点刻划清楚了。是不是呢?

[ 本帖最后由 老封 于 2007-5-17 08:50 编辑 ].

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当P点取成△DEF的外心、垂心等特殊点时,就可分别得到已经讨论过的那两种特例。换句话说,不动点P′在外接圆上的位置取决于P相对于△DEF的位置,两者是通过西摩松线来沟通的。.

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至于欧拉线所经过的定点,我也有了一个深入的结论:

结论四    △ABC的每边所在直线关于三边对称线(其中有一条不动)围成图中阴影的黄色三角形,共三个,其相应的欧拉线所围成的三角形记为△A′B′C′。任意直线L关于△ABC三边的对称直线所围成的三角形记为△D′E′F′。当L绕定点P旋转时,则△D′E′F′的欧拉线也必绕一个定点P′旋转,且P′、A′、B′、C′四点一定与P、A、B、C四点反向相似!

[ 本帖最后由 老封 于 2007-5-16 10:11 编辑 ].

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我来揭示这个图形的意义
其实研究一条直线关于三边对称线这一课题是对西摩松线的斯坦纳定理之深化。
斯坦纳定理指出:“外接圆上任一点关于三边的轴对称点一定共线,所共直线还经过△ABC的垂心。”
斯坦纳定理的逆定理是:“一条直线关于△ABC的轴对称线共点的充要条件是它通过垂心H,而且所共点一定落在外接圆上!”(参见梁绍鸿《初等数学复习及研究》复习题三第39题)
上述结论二是对这一逆定理的加强,给出了不共线情况的定量描绘。也就是说,三条轴对称线围成的三角形的大小取决于垂心H离开直线L的距离。
注:三对称直线所围成的三角形总与△ABC的垂三角形反向相似。.

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原来如此。.

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我正在写这题与三相似图形的内在联系,但刚刚来了客人,中断了。
网络又不好,忽断忽续,写了老半天,还没有整理成文呢.

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但我上面为什么说这个图形和轴对称的关系不密切呢?

原因是,可以从另一角度来理解这个图形:设作△ABC每一个顶点关于对边的轴对称点A′、B′、C′,得到四个相似三角形——△A′BC∽△AB′C∽△ABC′∽△ABC,其实△ABC与前三个是否相似并非问题之关键,只要周围三个三角形按这样的方式相似,就可把它们理解成三张相似的“地图”,问题就变成三张图中的对应直线何时共点这一颇有意味的问题了。

所以说,本图中的△ABC的外接圆其实就是三相似图形的“布洛卡(Brocard)圆”,一旦三对应线共点,所共点就一定落在该圆上!.

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关于一般的三相似图形

对于更一般的三相似图形,也就是说:将任意三张顺相似的“地图”重叠在同一平面上,这时可以提出一连串有意思的问题。例如:何时三对应点共线?何时三对应线共点?所共点形成的轨迹是什么?等等。这些都是带有一般意义的基本问题,值得讨论。而这正是《近代欧氏几何学》最后一章的研究范围。
只要三相似图形不退化,就一定能从中找到满足:△A′BC∽△AB′C∽△ABC′的六个点。其中,
A是地图Ⅱ和地图Ⅲ的相似不动点,A′是这个不动点在地图Ⅰ中的相应点;
B是地图Ⅲ和地图Ⅰ的相似不动点,B′是这个不动点在地图Ⅱ中的相应点;
C是地图Ⅰ和地图Ⅱ的相似不动点,C′是这个不动点在地图Ⅲ中的相应点。
这三个相似三角形的形状称为该三相似图形的“特征三角形”。
关于这种一般情形,其实有不少结论还是可以跟上面的特殊情形进行类比的。

[ 本帖最后由 老封 于 2007-5-16 14:29 编辑 ].

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刚收到美国数学奥林匹克国家队教练冯祖鸣先生邮件,里面提到一个平面几何题:

Hi, any interested problem can be made from this:
Let $ABC$ be a triangle, with incenter $I$. Its incircle touching the sides $BC, CA, AB$ at $D, E, F$, respectively. Line $EF$ intersects lines $BI$ and $CI$ at $Q$ and $P$, respectively. Lines $BP$ and $CQ$ meet at $X$. Then $XI$ is perpendicular to $BC$.
Any new geo. problems?Best


其实,这道题恰好等价于我为江苏省第十届初中数学竞赛(1995年?)所供的平面几何题:
“已知三条线段AB,BC,CD都与圆O相切,且AB=BC=CD,联结AC,BD交于X点,T是BC上的切点。求证:XT⊥BC。”

有兴趣的同学不妨思考一下,并比较一下两者的异同。

[ 本帖最后由 老封 于 2007-5-16 14:28 编辑 ].

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在一般的三相似图形中,可以保证AA′、BB′、CC′这三条联线一定共点,所共点P称为该三相似图形的“广义Fermat点”。
不难证明,三对应直线所围成的三角形a′b′c′的形状是固定的,它总是反向相似于P关于△ABC的垂足三角形abc。而且在△a′b′c′中找出P关于△abc的相似对应点,记其为P′。那么P′一定位于△ABC的外接圆——三相似图形的布洛卡圆上!
这就是命题一在三相似图形中的推广形式。在特殊情形中,H是垂三角形的内心,因此P′也就取内心了。

[ 本帖最后由 老封 于 2007-5-16 15:20 编辑 ].

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在一般三相似图形中,命题二也有其推广形式:
广义Fermat点关于△ABC的等角共轭点称为“广义等力点”,记其为Q,则Q关于△ABC的垂足三角形一定反向相似于特征三角形。

[ 本帖最后由 老封 于 2007-5-16 15:40 编辑 ].

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有意思的是:在特征三角形△A′BC、△AB′C、△ABC′中,与Q点对于其垂足三角形位置相应点Q1、Q2、Q3就是联线与AA′、BB′、CC′与布洛卡圆各自的交点!
与命题二相应的结论是:“三对应直线当且仅当各自经过Q1、Q2、Q3时共点,所共点也一定落在布洛卡圆上!”
而且,当三对应线并不共线时,它们所围成的三角形之大小,也与对应直线离开Q点的距离成正比。只是像命题二那样完美的定量结论对于一般的三相似图形目前还暂没有找到,有待高手们努力!.

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又一个新结论

结论五    当直线L绕着定点T旋转时,三条对称线所围成的三角形D′E′F′中的固定点P′的轨迹是圆,其半径r′等于
                                                             (PH×TQ)/2r,
其中P是垂三角形DEF中相应于P′的点,Q就是结论三中所构造的那个点,r是△DEF内切圆的半径。


结论三其实是结论五的特例,当r′退化为0时的情形。

[ 本帖最后由 老封 于 2007-5-17 09:56 编辑 ].

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三相似图形研究延续

关于三相似图形研究,有两项小进展:
(1)当三相似直线(在各自的地图中)绕某定点旋转时,所围成的三角形a′b′c′中的定点D′之轨迹总是圆;而且,不管旋转中心怎么移动,轨迹圆总与布洛卡圆交于一个固定的点X,这个X在布洛卡圆上的位置由D′在三角形a′b′c′中的位置所决定。设D′点在广义Fermat点的垂足三角形中的相应位置为D,则固定点X可如下刻划——它关于△ABC的西摩松线恰平行于直线PD!
这使结论三有了延续,但定量方面的结论还未最终找到。然而线索已经发现:像结论五那样,轨迹圆的大小仍取决于旋转中心离开特征三角形中一个神秘点的距离,这个神秘点相当于结论三中费了九牛二虎之力构造出来的那个Q点。不过在一般三相似图形中如何刻划它?这是个绕有趣味的问题。
(2)当三相似直线绕各自特征三角形的外接圆上某定点旋转时,所围成的三角形a′b′c′的外接圆必经过布洛卡圆上的一个定点。这两个定点间的联系亦尚未刻划成功。.

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出个题给大家思考:

“如图,设任意点关于△ABC三边的对称点分别为D、E、F。△ABC的外心和垂心分别为O、H。
求证: S△DEH∶S△DFH=S△ABO:S△ACO。”.

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不要小看这题,它在三相似图形中具有较深的涵义。.

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谁能给出这题的好的证法?
老封又将送出一本书作为悬赏。.

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我来提供一个新思路

这两天一直在思考这道涉及面积的命题,其实这个结论是我一个朋友于1999年前后就提出了,但我好像总没找到简单的证明。为了给出有效证明,近日我产生了一个推广斯坦纳(Steiner)定理的有趣思路:
当P点在△ABC的外接圆上时,由斯坦纳定理知:P关于AC、AB的对称点E、F与H三点共线。
当P点不在外接圆上时,发现△HEF的面积取决于P离开外心O的距离。我已推导了定量的表达式:.

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有了这个公式,那么前面那道涉及面积比的题目就不难证了,大家不妨试试。
而且,还可将垂足三角形面积的欧拉公式作为其推论:.

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2007-5-21 11:38

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斯坦纳,一位伟大的平面几何专家,他总是不倦地在思考一些问题。.

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2007-5-21 11:51

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一个深刻的结论,悬赏求解

这个问题已由安徽唐传发老师给出巧妙的证法!
今天早上他打电话来告诉了我他的绝妙构思,这样做来本题就并不是很难了,现暂且将此题撤去。

[ 本帖最后由 老封 于 2007-5-25 11:00 编辑 ].

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一位友好的美国数学家

前天晚上,数学奥林匹克今年国家队的副领队、广州大学计算机软件研究所所长朱华伟教授来电告诉说,他上周曾专赴加拿大,与台湾九章图书公司老板孙文先先生一起,代表中方接受了前美国数学会主席Klamkin先生遗赠给我国的2500余册数学原版书,并将这些图书运回到广州,整整摆放了两间房间!听说其中还包括不少珍贵的平面几何书呢。
Murray Seymour Klamkin,一位卓有成绩的数学家,1921年出生于纽约,2004年去世。他本人对几何十分专长,曾提出过著名的“Klamkin中线对偶原理”,揭示了由三角形的三条中线所围成的三角形与原三角形的一种内在联系,不仅它们的面积有一种明确的比例关系——3比4,而且,再取由第二个三角形的三条中线,所围成的三角形一定与原先的三角形相似,相似比也是3比4。这一原理可有效应用于几何不等式中。
这里留一个题给初中同学思考一下:
“已知一个三角形的三边之长与三条中线之长成比例,问这个三角形必须满足什么条件?”
除了正三角形这一平凡情形外,还有另外一种有趣情形。因为题中并未明确某条中线必需和哪一边成比例,于是出现了其它可能性。大家不妨一试。
注:本题取自毛鸿翔等《直线形》(江苏人民出版社,1980年版)一书的总复习题第35题。.

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2007-5-25 09:11

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下面这篇是华裔数学家Andy Liu所写的纪念Klamkin教授一文:

Memorial Celebration of the Life of Murray Seymour Klamkin   by Andy Liu


Let me first make it clear that this is not a eulogy. By my definition, a eulogy is an attempt to make the life of the departed sound much better than it was. In the present case, it is not only unnecessary, it is actually impossible. Murray Seymour Klamkin had a most productive and fulfilling life, divided between industry and academia.

Of the early part of his life, I knew little except that he was born in 1921 in Brooklyn, New York, where his father owned a bakery. This apparently induced in him his life-long fondness for bread. I read in his curriculum vita that his undergraduate degree in Chemical Engineering was obtained in 1942 from Cooper Union’s School of Engineering. During the war, he was attached to a chemical warfare unit stationed in Maryland, as his younger sister Mrs. Judith Horn informed me.

In 1947, Murray obtained a Master of Science degree from the Polytechnic Institute of New York, and taught there until 1957 when he joined AVCO’s Research and Advanced Development Division.

In 1962, Murray returned briefly to academia as a professor at SUNY, Buffalo, and then became a visiting professor at the University of Minnesota. In 1965, he felt again the lure of industry and joined Ford Motor Company as the Principal Research Scientist, staying there until 1976.

During all this time, Murray had been extremely active in the field of mathematics problem solving. His main contribution was serving as the editor of the problem section of SIAM Review. He had a close working relation with the Mathematical Association of America, partly arising from his involvement with the William Lowell Putnam Mathematics Competition.

In 1972, the MAA started the USA Mathematical Olympiad, paving the way for the country’s entry into the International Mathematical Olympiad in 1974, hosted by what was still East Germany.

Murray was unable to obtain from Ford release time to coach the team. Disappointed, he began to look elsewhere for an alternative career. This was what brought him to Canada, at first as a Professor of Applied Mathematics at the University of Waterloo.

However, it was not until the offer came from the University of Alberta that made up his mind to leave Ford. I did not know if Murray had been to Banff before, but he must have visited this tourist spot during the negotiation period, fell in love with the place and closed the deal.

As Chair, Murray brought with him a management style from the private sector. Apparently not everyone was happy with that, but he did light some fires under several pairs of pants, and rekindled the research programs of the wearers.

Murray had always been interested in Euclidean Geometry. He often told me about his high school years when he and a friend would challenge each other to perform various Euclidean constructions. Although the Chair had no teaching duties at the time, Murray took on a geometry class himself.

At the same time, Murray began editing the Olympiad Corner in Crux Mathematicorum, a magazine then published privately by Professor Leo Sauve of Ottawa. It is now an official journal of the Canadian Mathematical Society. Murray also introduced the Freshmen and Undergraduate Mathematics Competitions in the Department.

Geometry, mathematics competitions and Crux Mathematicorum were what brought me to Murray’s attention. At the time, I was a post-doctoral fellow seeking employment, having just graduated from his Department. Thus I was ready to do anything, and it happened that my interests coincided with those of Murray. I was holding office hours for his geometry class, helping to run the Department’s competitions and assisting him in his editorial duty.

I remember being called into his office one day. He had just received a problem proposal for Crux Mathematicorum. “Here is a nice problem,” he said, “but the proposer’s solution is crappy. Come up with a nice solution, and I need it by Friday afternoon!”

As much as I liked problem-solving, I was not sure that I could produce results by an industrial schedule. Nevertheless, I found that I did respond to challenges, and although I was not able to satisfy him every time, I managed to do much better than if I was left on my own, especially after I had got over the initial culture shock.

The late seventies were hard times for academics, with few openings in post-secondary institutions. I was short-listed for every position offered by the Department, but always came just short. Eventually, I went elsewhere for a year as sabbatical replacement. Murray came over to interview me for a new position, pushed my appointment through the Hiring Committee and brought me back in 1980.

Murray had been the Deputy Leader for the USA National Team in the IMO since 1975. In 1981, USA became the host of the event, held outside Europe for the first time. Sam Greitzer, the usual Leader, became the chief organizer. Murray took over as the Leader, and secured my appointment as his Deputy Leader.

I stayed in that position for four year, and in 1982, made my first trip to Europe because the IMO was in Budapest. This was followed by IMO 1983 in Paris, and IMO 1984 in Prague. I was overawed by the international assembly, but found that they in turn were overawed by Murray’s presence. He was arguably the most well-known mathematics problem-solver in the whole world.

We both retired from the IMO after 1984, even though I would later return to it. His term as Chair also expired in 1981. Thus our relationship became collegial and personal. He and his wife Irene had no children, but they were very fond of company. I found myself a guest at their place at regular interval, and they visited my humble abode a few times.

It was during this period that I saw a different side of Murray. Before, I found him very businesslike, his immense talent shining through his incisive insight and clinical efficiency.

Now I found him a warm person with many diverse interest, including classical music, ballroom dancing, adventure novels, kung-fu movies and sports, in particular basketball.

Although Murray had been highly successful in everything he attempted, he will probably be remembered the most for his involvement in mathematics problem-solving and competitions. He had authored or edited four problem books, and had left his mark in every major journal which had a problem section. He had received an Honorary Doctorate from the University of Waterloo and was a Fellow of the Royal Society of Belgium. He had won numerous prizes, and had some named after him.

Murray had enjoyed remarkably good health during his long life. It began to deteriorate in September 2000 when he underwent a by-pass operation. After his release from the hospital, he continued to exert himself, walking up to his office on the sixth floor, and skating in the West Edmonton Mall.

His heart valve gave in November, fortunately while he was already in the hospital for physiotherapy. He was in coma for some time. One day, when I visited him, he was bleeding profusely from his aorta. The doctor indicated to me that he did not expect Murray to last through the day.

Somehow, the inner strength of Murray came through, and on my next visit, he was fully conscious. He told me to make arrangement for his eightieth birthday party, stating simply that he would be out of the hospital by that time. It was a good thing that I took his words seriously, for he was out of the hospital by that time, ready to celebrate.

One of the last mathematical commitment he made was to edit the problem section in the MAA’s new journal Math Horizons. During this difficult time, he asked me to serve with him as joint-editors. Later, he passed the column onto me, but his finger-prints were still all over the pages.

Now I have to try to fill in his shoes without the benefit of his wisdom. His passing marks the end of an era in the world of mathematics competition and problem-solving. He will be deeply missed..

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这里改换一题替代原先的悬赏题:

“如图,A、B、C是平面上任意三点。在两圆⊙O1、⊙O2上分别有四个点A1、A2、A3、A4;B1、B2、B3、B4,满足四边形A1A2A3A4与B1B2B3B4顺相似。然后作△C1AB∽△CA1B1,△C2AB∽△CA2B2,△C3AB∽△CA3B3,△C4AB∽△CA4B4。求证:C1、C2、C3、C4四点共圆!”

注:这也是我在1999年时与一位朋友合作的一项研究成果,可惜那位似乎已对数学不感兴趣了。.

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老封的问题都比较深刻,这段时间忙于它务,也没有时间思考,看来只有等待暑假了。
我这里有个学生问题,已有证明,但我没看出它的出题背景,请老封和各位大虾赐教:.

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2007-5-28 07:29

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回复 #143 老殿 的帖子

这个题目我思考过,有一定心得。下次慢慢细谈.

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最有意思的是,我用《几何专家》证明这个问题时出现一个问题:用GDD方法证明,超时;用面积方法证明结论是命题错误!用吴法证明结论是命题正确,用向量方法证明结论又是命题错误! 你说怪不怪?看来机器定理证明的完善还有很长的道路要走。.

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回复 #145 老殿 的帖子

呵呵,电脑在作怪啊。有时电脑还不如人脑好使呢.

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回复 #143 老殿 的帖子

我记得这道题的原型是BF为∠ABC的外交平分线吧,这一样的,前一段时间好像刚刚做过.

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回复 #147 Schumi 的帖子

这道题的原型是BF为∠ABC的外角平分线,上面的是几何专家里的作图法,为了照顾证明(尽量使图形简单,另一个原因是几何专家中竟然没有角平分线作图!——我认为这也是限制几何专家推广的一个重要原因:为了照顾证明,不能自由绘图探索几何性质)修改而得的。.

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回复 #146 老封 的帖子

应该不是电脑的毛病,还应该是软件设计不够完善所致。.

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我来点评老殿提到的题

这是一道较常见的几何习题,例如较早时就见于《数学题解》(吉林人民出版社1980年3月版,孙诲正 王得福 于永泉编)一书的(下册)第338页第1题。
如运用初三的知识,可直接证D、E、B、F四点共圆,极为方便;如只运用初二全等三角形的知识,则需如图添一条辅助线EG,然后证△DGE≌△EBF。.

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