[Mathematics from the beginning] No. 222 quarter with a computer to the college Entrance Examination (14)

plot summary:
[Machine Xiao Wei] in [engineer Ah Wei] escorted into the [nine turn elixir] of the turn of the practice. Imagine a scenario:
If you were allowed to take an Internet-connected computer to the college entrance exam, would you give up choosing a hand-held calculator and a draft book?
？ Ah Wei decided and Xiao Wei to try to use the computer to calculate high questions will be how the feeling.

Drama Start:

Star Calendar May 25, 2016 17:14:22, the Milky Way Galaxy Earles the Chinese Empire Jiangnan Line province.
[Engineer Ah Wei] is working with [machine Xiao Wei] to do 2013 years of Jiangsu Province Mathematics college entrance exam questions].

The above sentence is really not true in this article, because this paper Ah Wei and [machine Xiao Wei]

No one has done it, purely punchy.

A cursory look at, the difficulty is still 5.5 ring, that is, belong to the difficult ranks. Imagine every time you

In order to get a piece of treasure have to be 5.5 times the feeling of flirting, that is.

That year Ah Wei never dared to hand over blank, now gall also fat, also dare has failed, really very dashing feeling.

This time, Ah Wei and [machine Xiao Wei] mainly in the study of how to solve the two Yuan two equations, first look at this test question:

These two curves are this diagram:

The reason why we can solve the two-yuan two-time equations is that they satisfy such relationships in their coefficients:

<span style= "FONT-SIZE:18PX;" >>>> [' (0) ', ' (2) *b_[1]^[1]*b_[2]^[1] ', ' (0) ', ' (4) *c_[1]^[1] ', ' (4) *c_[2]^[1] ' [' (4) *b_[1]^[2]*b_[2]^[ 2] ', ' ( -16) *b_[1]^[2]*c_[2]^[1] ', ' ( -16) *b_[2]^[2]*c_[1]^[1] ', ' (+) *c_[1]^[1]*c_[2]^[1] ' [' (0) ', ' (0) ', ' (0) ', ' (0) ', ' (4) *b_[1]^[2]*b_[2]^[2] ', ' (*b_[1]^[1]*b_[2]^[1]*c_[1]^[1) ', ' (+) *b_[1]^[1]*b_[2]^[1]*c_[2]^[1] ', ' (+) *c_ [1]^[2] ', ' (+) *c_[1]^[1]*c_[2]^[1] ', ' (+) *c_[2]^[2] ' [' (0) ', ' (0) ', ' (+) *b_[1]^[1]*b_[2]^[1]*c_[1]^[1] ', ' (*b) _[1]^[1]*b_[2]^[1]*c_[2]^[1] ', ' (+) *c_[1]^[2] ', ' ( -32) *c_[1]^[1]*c_[2]^[1] ', ' (+) *c_[2]^[2] ', ' (*B_[1]^[2]*C_) [2]^[1] ', ' (+) *b_[2]^[2]*c_[1]^[1] '] #测试, two two the identity of the root of the equation set, has eliminated a dollar # first to ensure that the two equation of the elimination of the highest sub-coefficient of the element is 1def Tmp3 (): Part1 =    Alg.strformat ([' b_[1] ', ' b_[2] ');    Part2 = Alg.strformat ([' b_[1]^[2] ', ' -4c_[1] ');    Part3 = Alg.strformat ([' b_[2]^[2] ', ' -4c_[2] ');    Part4 = Alg.strpow_n (part1, 2);    Part5 = Alg.stradd (Part2, PART3);    part6= Alg.strcombine (Alg.stradd (Part4, Alg.minus (PART5))); #print (PART6);    Part7 = Alg.strdot (Alg.strdot ([' (4) '], part2), part3);    Part7 = Alg.strcombine (PART7);    #print (PART7);    PART8 = Alg.strpow_n (part6, 2);    PART8 = Alg.strcombine (PART8);    #print (PART8);    result = Alg.strcombine (Alg.stradd (PART8, Alg.minus (PART7))); Print (Result);>>> [' (0.0) ', ' (0.0) ', ' (1.0) *b_[1]^[1]*b_[2]^[1]*c_[1]^[1] ', ' (1.0) *B_[1]^[1]*B_[2]^[1]*C_ [2]^[1] ', ' (1.0) *c_[1]^[2] ', ' ( -2.0) *c_[1]^[1]*c_[2]^[1] ', ' (1.0) *c_[2]^[2] ', ' (1.0) *b_[1]^[2]*c_[2]^[1] ', ' (1.0) *b _[2]^[2]*C_[1]^[1] ']</span>

<span style= "FONT-SIZE:18PX;" >var s = [' in a_[1] = = a_[2] = = 1 O'Clock ', ' ((B_[1]-b_[2]) ^[2]-(b_[1]^[2]-4c_[1]+b_[2]^[2]-4c_[2])) ^[2] ', ' = = 4 (b_[1]^[2]-4c_ [1]) (b_[2]^[2]-4c_[2]) ', ', ' This identity can be translated into: ', ' (1) *b_[1]*b_[2]*c_[1]+ (1) *b_[1]*b_[2]*c_[2] ', ' + (1) *c_[1]^[2]-(2) *c_[1]*c_ [2]+ (1) *c_[2]^[2] ', ' + (1) *b_[1]^[2]*c_[2]+ (1) *b_[2]^[2]*c_[1] ', ' = 0 '];</span>

Add two new methods to the tool:

<span style= "FONT-SIZE:18PX;" > #把一个只包括 + number polynomial string split into a polynomial array #如 ' (1) *x^[2]+ ( -1) ' = = [' (1) *x^[2] ', ' ( -1) '] def str2array (self, str1): arr        ay = [];        #加号位置 Signindex = str1.find (' + ');        Print (Signindex);        start = 0;        Count = 0; while (signindex! = '-1 ' and Count <): #符合要求的必须连着下一个单项式的系数 #按照统一格式是左括号开始 if Str1[si                GNINDEX+1] = = ' (': Array.append (Str1[start:signindex]);            Start = Signindex + 1;            Signindex = str1.find (' + ', signindex+1);        if (Signindex = =-1): break;        Array.append (Str1[start:]); return array;>>> 9[' (1) *x^[2] ', ' ( -1) '] #解二元二次方程组 def solveEquationExp2 (self, Array1, array2): #输入的是 Two coefficient matrices #矩阵具有这样的形式: [[' 1 '], [b_[1]], [c_[1]]] #也就是对于ax ^[2]+bx+c=0, a=1, and B, C is an array #注意 with a parametric polynomial, the coefficients are arrays, not characters        Strings and so on.        A_1, b_1, c_1 = Array1[0], array1[1], array1[2]; A_2, b_2, c_2 = array2[0], arrAY2[1], array2[2];        #恒等式 ' [' (1.0) *b_[1]^[1]*b_[2]^[1]*c_[1]^[1] ', ' (1.0) *b_[1]^[1]*b_[2]^[1]*c_[2]^[1] ',        ' (1.0) *c_[1]^[2] ', ' ( -2.0) *c_[1]^[1]*c_[2]^[1] ', ' (1.0) *c_[2]^[2] ', ' (1.0) *b_[1]^[2]*c_[2]^[1] ',        ' (1.0) *b_[2]^[2]*c_[1]^[1] ' = 0 "#这些运算都是针对数组的 b1b2 = Alg.strcombine (Alg.strdot (b_1, b_2));        C1C2 = Alg.strcombine (Alg.strdot (c_1, c_2));        C12 = Alg.strcombine (Alg.strpow_n (c_1, 2));        C22 = Alg.strcombine (Alg.strpow_n (c_2, 2));        B12 = Alg.strcombine (Alg.strpow_n (b_1, 2));                B22 = Alg.strcombine (Alg.strpow_n (b_2, 2));        Part1 = Alg.strcombine (Alg.strdot (b1b2, Alg.stradd (c_1, c_2)));        Part2 = Alg.strcombine (C12+alg.strdot ([' ( -2) '], c1c2) +C22);        Part3 = Alg.strcombine (Alg.stradd (Alg.strdot (B12, c_2), Alg.strdot (B22, c_1)));        result = Alg.strcombine (PART1+PART2+PART3); Return RESULT;&LT;/SPAn> 

This makes it possible to solve the problem in a routine way:

<span style= "FONT-SIZE:18PX;" >>>> Step1: [' (1) *x^[2] ', ' (1) *y^[2] ', ' (2) *x*y ', ' ( -1) ']step2: [' (1) ', ' (2) *x ', ' (1) *x^[2]+ ( -1) ']step1: [' (1) *x^[2] ', ' (4) *y^[2] ', ' ( -1) ']step2: [' (4) ', ' 0 ', ' (1) *x^[2]+ ( -1) ']step3: [' (1) '] [' (2) *x '] [' (1) *x^[2] ', ' ( -1) ']step  3: [' (4) '] [' 0 '] [' (1] *x^[2] ', ' ( -1) ']step4: [' (0) ', ' (4) *x^[4] ', ' ( -4) *x^[2] ']STEP5: [4, 0, -4, 0, 0]step6: [1.-1. 0.0.] STEP7: [' (((2) *x)) + ((((2) (*x) ^[2]) + ((( -4) * ((1) * ((1) *x^[2]+ ( -1))))/((2) * ((1))) ', ' ((((-(-((2) *x))-((( ((2) *x) ^[2]) + (( -4) * ((1) * ((1) *x^[2]+ ( -1)))/((2) * ((1)) ']STEP8: [[1.0], 0.0], [1.0,-2.0], [0.0, 1.0], [0.0],    -1.0], [0.0, 1.0], [0.0, -1.0]] #测试def tmp2 (): Solve = Stringalgsolve ();    f = Alg.strformat ([' x^[2] ', ' y^[2] ', ' 2xy ', '-1 ']);    Print (' Step1: ', f);    Poly_y_f = Solve.coefarray (f, ' Y ');    Print (' Step2: ', poly_y_f);    #以y为参数的二次多项式的系数, the extinction is A1 y, b1, C1 = Poly_y_f[0], poly_y_f[1], poly_y_f[2];    g = Alg.strformat ([' x^[2] ', ' 4y^[2] ', '-1 ']); Print ('Step1: ', g);    Poly_y_g = Solve.coefarray (g, ' Y ');    Print (' Step2: ', poly_y_g);        #以y为参数的二次多项式的系数, the extinction is A2 y, b2, c2 = poly_y_g[0], poly_y_g[1], poly_y_g[2]; A1, B1, C1 =solve.str2array (A1), Solve.str2array (B1), Solve.str2array (C1) A2, B2, C2 =solve.str2array (A2), Solve.str2arr    Ay (B2), Solve.str2array (C2) print (' Step3: ', A1, B1, C1);    Print (' Step3: ', A2, B2, C2);    poly_x = SOLVE.SOLVEEQUATIONEXP2 ([A1, B1, C1], [A2, B2, C2]);    Print (' Step4: ', poly_x);    poly_x = Solve.coefpoly (poly_x, ' X ');    Print (' Step5: ', poly_x);    #求得x的根 roots = Np.roots (poly_x);    Print (' Step6: ', Roots);    #求方程式 <1> y expression for x expr_y_root = Solve.solvepoly (Poly_y_f);    Print (' Step7: ', expr_y_root);    Expr_y_root2 = Solve.solvepoly (poly_y_g);        Print (' Step7: ', Expr_y_root2);    #求相交点的坐标对组 points = [];    Points2 = [];        For I in range (len (roots)): Real = ABS (roots[i].real);        Abs_ = ABS (Roots[i]);     #实数根 if ABS (Real-abs_) < 0.001:                   For j in Range (Len (expr_y_root)): x = Roots[i];                y = Solve.streval (expr_y_root[j], ' x ', x);                Points.append ([x, y]);                y = Solve.streval (expr_y_root2[j], ' x ', x);    Points2.append ([x, y]);    Print (' Step8: ', points);    Print (' Step8: ', points2); #比较两组点, draw the intersection [1, 0], [-1, 0]</span>

Solution to this end, the following are some of the process artifacts, can be ignored.

<span style= "FONT-SIZE:18PX;"                    >if (1) {var r = 20;                      Config.setsector (1,1,1,1);                    Config.graphpaper2d (0, 0, R);                                      Config.axis2d (0, 0,180);                  Axis set var ScaleX = 2*r, ScaleY = 2*r;                   var SpaceX = 0.4, SpaceY = 0.4;                  var XS = -10, XE = 10;                  var YS = -10, YE = 10;                    Config.axisspacing (XS, XE, SpaceX, ScaleX, ' X ');                                        Config.axisspacing (YS, YE, SpaceY, ScaleY, ' Y ');                    var transform = new transform ();                                    Store the points on the function image var a = [], b = [], c = [], d = []; Function description//Greek alphabet to be displayed (save this for ctrl c/v//αβγδεζηθικλμνξ0πρστυφχψω//αβγδεζηθικλ                              Μνξοπρστυφχψωvar f1 = ' X+y = 1 ', F2 = ' X+y=-1 ', F3 = ' x^[2]+4y^[2] = 1 ', F4 = '; //function stroke//parametric equation var x, y;                    var pointa = [];            for (var Thita = 0; Thita < math.pi*2; Thita +=math.pi/48) {x = Math.Cos (Thita); A.push ([X, 1-x]); B.push ([X, -1-x]);          C.push ([Math.Cos (Thita), 0.5*math.sin (Thita)]);                                  }//Store temporary array var tmp = []; Show transform if (A.length > 0) {a = Transform.scale (Transform.translate (A, 0, 0), scalex/s                     Pacex, Scaley/spacey);                        function 1 TMP = [].concat (a);                        Shape.pointdraw (TMP, ' red ');                        TMP = [].concat (a);                                            Shape.multilinedraw (tmp, ' Pink ');                    Plot.setfillstyle (' Red ');                  Plot.filltext (F1, 100,-90, 200); }//Show transform if (B.length > 0) {b = Transform.scale (transform.translate (b, 0,0), Scalex/spacex, Scaley/spacey);                        function 1 TMP = [].concat (b);                        Shape.pointdraw (TMP, ' red ');                        TMP = [].concat (b);                                            Shape.multilinedraw (TMP, ' green ');                    Plot.setfillstyle (' green ');                  Plot.filltext (F2, 100,-120, 200); }//Show transform if (C.length > 0) {c = Transform.scale (Transform.translate (c, 0, 0), scale                     X/spacex, Scaley/spacey);                        function 1 TMP = [].concat (c);                        Shape.pointdraw (tmp, ' Blue ');                        TMP = [].concat (c);                                            Shape.multilinedraw (tmp, ' #0088FF ');                    Plot.setfillstyle (' Blue ');                  Plot.filltext (F3, 100,-150, 200); }}</span>

<span style= "FONT-SIZE:18PX;"  >//test if (1) {var mathtext = new Mathtext ();//Greek alphabet (save this for ctrl c/v//αβγδεζηθικλμνξ0πρστυφχψω//αβγδεζηθικλμνξοπρστυ φχψω//Greek uppercase and lowercase letter var Gc = ' αβγδεζηθικλμνξ0πρστυφχψω '; var Gs = ' αβγδεζηθικλμνξοπρστυφχψω '; var s = [/* ' Test: Seek intersection ', ' x^[2]+2xy+y^ [2]=1 _[(1)] ', ' x^[2]+4y^[2] = 1  _[(2)] ', ', ' Step1: Polynomial of two curves ', ' step2: Elimination of Y ', ' step3: two sets of two coefficients into equation identity ', ' Step4: Get equation identity , this is the four-time identity of meta-X, without the radical ', ' step5: the coefficient array of the derived identity. ', */' STEP6: The root of the Jie Heng equation has four roots. ', ' STEP7: The algebraic formula of the Y-element's root in the curve equation is obtained. ', ' STEP8: The algebraic formula is evaluated to get the intersection coordinates '];</span>

From then on, [machine Xiao Wei] can be vertical and horizontal conic.

Leave a snapshot:

Now Ah Wei really cool, feel the outside scenery is also beautiful up.

The end of this section, to know how to funeral, please see tell.

Note: Although the solution in this section is correct, the final result is wrong, because the identity is not strong enough,

Due to the lack of time, although Ah Wei realized that the solution is wrong, there is no way, the next section will give the correct solution

As well as the results, everyone can go to see. Of course, knowing that this solution is not getting the right result is also a harvest.

[Mathematics from the beginning] No. 222 quarter with a computer to the college Entrance Examination (14)