### 数学专业英语－Mathematical Discovery

<P ><FONT face="Times New Roman" size=3>To give the flavor of Polya’s thinking and writing in a very beautiful but subtle case , a case that involve a change in the conceptual mode , I shall quote at length from his Mathematical Discovery (vol.II , pp.54 ff): </FONT></P><P ><FONT face="Times New Roman"><FONT size=3> EXAMPLE I take the liberty a little experiment with the reader , I shall state a simple but not too commonplace theorem of geometry , and then I shall try to reconstruct the sequence of idoas that led to its proof . I shall proceed slowly , very slowly , revealing one clue after the other , and revealing each gradually . I think that before I have finished the whole story , the reader will seize the main idea (unless there is some special hampering circumstance ) . But this main idea is rather unexpected , and so the reader may experience the pleasure of a little discovery .</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> A.If three circles having the same radius pass through a point , the circle through their other three points of intersection also has the same radius .</FONT></FONT></P>

<Palign=center><v:shapetype><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path connecttype="rect" gradientshapeok="t" extrusionok="f"></v:path><lock aspectratio="t" v:ext="edit"></lock></v:shapetype><v:shape><v:imagedata><FONT face="Times New Roman" size=3></FONT></v:imagedata></v:shape></P>

<Palign=center><FONT face="Times New Roman" size=3>Fig.1 Three circles through one point.</FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> <p></p></FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> This is the theorem that we have to prove . The statement is short and clear , but does not show the details distinctly enough . If we draw a figure (Fig .1)and introduce suitable notation , we arrive at the following more explicit restatement :</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> B . Three circles k , l , m have the same radius r and pass through the same point O . Moreover , l and m intersect in the point A , m and k in B , k and l in C . Then the circle e through A , B , C has also the radius</FONT></FONT></P>

<Palign=center><v:shape><v:imagedata><FONT face="Times New Roman" size=3></FONT></v:imagedata></v:shape></P>

<Palign=center><FONT face="Times New Roman" size=3>Fig .2 too crowded .</FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> Fig .1 exhibits the four circles k , l , m , and e and their four points of intersection A, B , C , and O . The figure apt to be unsatisfactory , however , for it is not simple , and it is still incomplete ; something seems to be missing ; we failed to take into account something essential , it seems .</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> We are dialing with circles . What is a circle ? A circle is determined by center and radius ; all its points have the same distance , measured by the length of the radius , from the center . We failed to introduce the common radius r , and so we failed to take into account an essential part of the hypothesis . Let us , therefore , introduce the centers , K of k , L of l , and M of m . Where should we exhibit the radius r ? there seems to be no reason to treat any one of the three given circles k ; l , and m or any one of the three points of intersection A , B , and C better than the others . We are prompted to connect all three centers with all the points of intersection of the respective circle ; K with B , C , and O , and so forth .</FONT></FONT></P>

<P ><FONT size=3><FONT face="Times New Roman">The resulting figure (Fig . 2) is disconcertingly crowded . There are so many lines , straight and circular , that we have much trouble old</FONT>－<FONT face="Times New Roman">fashioned magazines . The drawing is ambiguous on purpose ; it presents a certain figure if you look t it in the usual way , but if you turn it to a certain position and look at it in a certain peculiar way , suddenly another figure flashes on you , suggesting some more or less witty comment on the first . Can you recognize in our puzzling figure , overladen with straight and circles , a second figure that makes sense ?</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3>We may hit in a flash on the right figure hidden in our overladen drawing , or we may recognize it gradually . We may be led to it by the effort to solve the proposed problem , or by some secondary , unessential circumstance . For instance , when we are about to redraw our unsatisfactory figure , we may observe that the whole figure is determined by its rectilinear part (Fig . 3) .</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> This observation seems to be significant . It certainly simplifies the geometric picture , and it possibly improves the logical situation . It leads us to restate our theorem in the following form .</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3>C . If the nine segments </FONT></FONT></P>

<Palign=center><FONT face="Times New Roman" size=3>KO , KC , KB ,</FONT></P>

<Palign=center><FONT face="Times New Roman" size=3>LC , LO , LA ,</FONT></P>

<Palign=center><FONT face="Times New Roman" size=3>MB , MA , MO ,</FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> <p></p></FONT></FONT></P>

<P ><FONT face="Times New Roman" size=3>are all equal to r , there exists a point E such that the three segments </FONT></P>

<Palign=center><FONT face="Times New Roman" size=3>EA , EB , EC ,</FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> <p></p></FONT></FONT></P>

<P ><FONT face="Times New Roman" size=3>are also equal to r .</FONT></P>

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<Palign=center><FONT size=3><FONT face="Times New Roman">Fig . 3 It reminds you </FONT>－<FONT face="Times New Roman">of what ?</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> This statement directs our attention to Fig . 3 . This figure is attractive ; it reminds us of something familiar . (Of what ?)</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> Of course , certain quadrilaterals in Fig .3 . such as OLAM have , by hypothesis , four equal sided , they are rhombi , A rhombus I a familiar object ; having recognized it , we can “see “ the figure better . (Of what does the whole figure remind us ?)</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3>Oppositc sides of a rhombus are parallel . Insisting on this remark , we realize that the 9 segments of Fig . 3 . are of three kinds ; segments of the same kind , such as AL , MO , and BK , are parallel to each other . (Of what does the figure remind us now ?)</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3>We should not forget the conclusion that we are required to attain . Let us assume that the conclusion is true . Introducing into the figure the center E or the circle e , and its three radii ending in A , B , and C , we obtain (supposedly ) still more rhombi , still more parallel segments ; see Fig . 4 . (Of what does the whole figure remind us now ?)</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3>Of course , Fig . 4 . is the projection of the 12 edges of a parallele piped having the particularity that the projection of all edges are of equal length . </FONT></FONT></P>

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<Palign=center><FONT face="Times New Roman" size=3>Fig . 4 of course !</FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> <p></p></FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3>Fig . 3 . is the projection of a “nontransparent “ parallelepiped ; we see only 3 faces , 7 vertices , and 9 edges ; 3 faces , 1 vertex , and 3 edges are invisible in this figure . Fig . 3 is just a part of Fig . 4 . but this part defines the whole figure . If the parallelepiped and the direction of projection are so chosen that the projections of the 9 edges represented in Fig . 3 are all equal to r (as they should be , by hypothesis ) , the projections of the 3 remaining edges must be equal to r . These 3 lines of length r are issued from the projection of the 8<SUP>th</SUP> , the invisible vertex , and this projection E is the center of a circle passing through the points A , B , and C , the radius of which is r .</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> Our theorem is proved , and proved by a surprising , artistic conception of a plane figure as the projection of a solid . (The proof uses notions of solid geometry . I hope that this is not a treat wrong , but if so it is easily redressed . Now that we can characterize the situation of the center E so simply , it is easy to examine the lengths EA , EB , and EC independently of any solid geometry . Yet we shall not insist on this point here .)</FONT></FONT></P>

<P ><FONT face="Times New Roman"><FONT size=3> This is very beautiful , but one wonders . Is this the “ light that breaks forth like the morning . “ the flash in which desire is fulfilled ? Or is it merely the wisdom of the Monday morning quarterback ? Do these ideas work out in the classroom ? Followups of attempts to reduce Polya’s program to practical pedagogics are difficult to interpret . There is more to teaching , apparently , than a good idea from a master .</FONT></FONT></P>

<Palign=right><B><FONT size=3>——<FONT face="Times New Roman">From Mathematical Experience <p></p></FONT></FONT></B></P>

<P ><FONT size=3><FONT face="Times New Roman"> <p></p></FONT></FONT></P> <DIV class=Section1 style="LAYOUT-GRID:15.6pt none"><P 0cm 0cm 0pt; TEXT-ALIGN: center" align=center><B><FONT face="Times New Roman">Vocabulary</FONT></B></P><P 0cm 0cm 0pt"><FONT face="Times New Roman"> <p></p></FONT></P></DIV><BR auto; mso-break-type: section-break" clear=all><DIV class=Section2 style="LAYOUT-GRID:15.6pt none"><P 0cm 0cm 0pt"><FONT face="Times New Roman">subtle </FONT>巧妙的，精细的</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">clue </FONT>线索，端倪</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">hamper </FONT>束缚，妨碍</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">disconcert </FONT>使混乱，使狼狈</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">ambiguous </FONT>含糊的，双关的</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">witty </FONT>多智的，有启发的</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">rhombi </FONT>菱形（复数）</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">rhombus </FONT>菱形</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">parallelepiped </FONT>平行六面体</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">projection </FONT>射影</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">solid geometry </FONT>立体几何</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">pedagogics </FONT>教育学，教授法</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">commonplace </FONT>老生常谈；平凡的</P></DIV><BR auto; mso-break-type: section-break" clear=all><DIV class=Section3 style="LAYOUT-GRID:15.6pt none"></DIV><BR always; mso-break-type: section-break" clear=all>

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