Wolfram言語による創造的な数学教育〜高校生がどのようにして数学公式を発見したか
-Creative Mathematics Education Using Wolfram Language-

runnerskg@gmail.com

アメリカ、イングランド、イタリア、オーストラリア、セルビア、韓国、日本などの数学系の雑誌で査読論文を出し、アメリカ、イングランド、韓国、シンガポール、マレーシア、フィリピン、日本などで研究発表を行ってきました。そのノウハウを伝えたいと思います。

We have published  refereed papers in math magazines in the U.S., England, Italy, Australia, Serbia, Korea and Japan. We also talked at conferences in the U.S., Singapore, Malaysia, Philippines and Japan. Here, I would like to talk about the method we used.

私と生徒が出した論文の一部 （査読論文数は30を超える）

1. Miyadera, R., Inoue, M. & Fukui, M.(2016).Impartial Chocolate Bar Games with a Pass, Integers, 16, G5.(アメリカ)
2. Naito, M., Yamauchi, T., Minematsu, D.and Miyadera, R. : The Josephus problem in both directions, Visual Mathematics Art and Science Electric Journal of ISIS - Symmetry, Volume series 12, No.series 3, (2010).(セルビア)
3. Matsui, H., Minematsu, D., Yamauchi, T. & Miyadera, R.(2010), Pascal - like triangles and Fibonacci - like sequences, Mathematical Gazette, 94 (529), (イギリス)
4. Miyadera, R., Hashiba, S., Minematsu, D. : How high school students can discover original ideas of mathematics using Mathematics.Mathematica in education and research series 11 (3) (2006).(オーストラリア)
5. Miyadera, R., Kotera, Y. : Una Bella Curva Che Troviamo in Connessione con lo Sviluppo di, Archimede, series 2 (2005).(イタリア)
また、the U.S., Singapore, Malaysia, Philippines and Japanでの国際学会での発表はこれまで１１回ある。

Mathematical Gazette (England) and Integers (U.S.)

My students with Prof. Erik D. Demaine (MIT) and Prof. Jin Akiyama at JCDCG^3.

My students with Mr. C. Federighi (Apple CTO)   and Mr. T. Cook (Apple CEO) at WWDC.

何故創造性を育てる教育が必要なのでしょう？

Why we need to cultivate creativity in education?

答は、この世界の中にある多くの難題を解決するには、創造的な人々を多く育てる必要があるからです。

The answer to this question is ...

Since we have a lot of difficult problems in the world and we need a lot of creative people who solve them.

創造性を育てることは難しいと考えられている。どうすればよいのか？

Many people think it difficult to cultivate creativity. What shall we do?

私は創造性を育てることは難しくないと考えます。その方法は、生徒達に創造的なことをしてもらうことです。

My idea is simple:
“The best way to teach creativity is to give students a chance to create something interesting.”

ではどのようにすれば創造的なことができるのでしょうか？

How can students be creative?

例　　正方形の紙でつくるコップ

Example.   A cup made by a Square Sheet of paper.

zukei2Db[x_]:=Block[{xx2},xx2[1]={6-x,6-x};xx2[2]={6-x,-6+x};xx2[3]={-6+x,-6+x};xx2[4]={-6+x,6-x};pts0={{6-x,6},{5-x,6-x/2},{6-x,6-x}};pts1={{6-x,6},{6-x+0/2,6-x/4},{6-x+0,6}};pts2={{6-x,6},{6-x-1,0},{6-x,-6}};
Graphics[{{Gray,Polygon[{{6,6},{5.99,6},{5.99,6}}]},Polygon[{{-6,6},{-5.99,6},{-5.99,6}}],Polygon[{{6,-6},{5.99,-6},{5.99,-6}}],Polygon[{{-6,-6},{-5.99,-6},{-5.99,-6}}],
Polygon[{{6,6},{5.99,6},{5.99,6}}],Polygon[{{-6,6},{-5.99,6},{-5.99,6}}],Polygon[{{6,-6},{5.99,-6},{5.99,-6}}],Polygon[{{-6,-6},{-5.99,-6},{-5.99,-6}}],{RGBColor[0.6,0.9,0.9],Polygon[{xx2[1],xx2[2],xx2[3],xx2[4]}]},{RGBColor[0.6,0.9,0.9],Polygon[{xx2[1],xx2[2],xx2[2]+{x,-0},xx2[1]+{x,0}}]},
{RGBColor[0.6,0.9,0.9],Polygon[{xx2[4],xx2[1],xx2[1]+{0,x},xx2[4]+{-0,x}}]},
{RGBColor[0.6,0.9,0.9],Polygon[{xx2[3],xx2[4],xx2[4]+{-x,0},xx2[3]+{-x,-0}}]},{RGBColor[0.6,0.9,0.9],Polygon[{xx2[2],xx2[3],xx2[3]+{-0,-x},xx2[2]+{0,-x}}]},{GrayLevel[0.5],BSplineCurve[pts2]},{GrayLevel[0.5],BSplineCurve[pts0]},{GrayLevel[0.5],BSplineCurve[pts1]},{Black,Text["xcm",{5-x,6-x/2}]},If[0>0.1,{Black,Text["ycm",{6-x+0/3,6-x/4}]},{RGBColor[0.6,0.9,0.9],Text["0cm",{0,0}]}],{Black,Text["12cm",{6-x-1,0}]}},Background->Black,AspectRatio->Automatic]];rotationmat[dg_,point1_,point2_,rotatepoint_]:=Block[{degr=dg,p1=point1,p2=point2,p3=rotatepoint},(RotationMatrix[dg,p1-p2].(p3-p1))+p1];zukei3Db[x_,α_,gr_]:=Block[{xx,h},xx[1]={6-x,6-x,0};xx[2]={6-x,-6+x,0};xx[3]={-6+x,-6+x,0};xx[4]={-6+x,6-x,0};h=(8/6) ((((x-0))^2)+(x-0) *x-18(x-0)-18 x+(x^2)+108)(Sqrt[(x^2)-(((0))^2)])//N;
Graphics3D[{If[gr==1,{White,Text[h,{3,3,6}]},{Black,Text[h,{3,3,6}]}],Polygon[{{6,6,0},{5.99,6,0},{5.99,6,0}}],Polygon[{{-6,6,0},{-5.99,6,0},{-5.99,6,0}}],Polygon[{{6,-6,0},{5.99,-6,0},{5.99,-6,0}}],Polygon[{{-6,-6,0},{-5.99,-6,0},{-5.99,-6,0}}],
Polygon[{{6,6,6},{5.99,6,6},{5.99,6,6}}],Polygon[{{-6,6,6},{-5.99,6,6},{-5.99,6,6}}],Polygon[{{6,-6,6},{5.99,-6,6},{5.99,-6,6}}],Polygon[{{-6,-6,6},{-5.99,-6,6},{-5.99,-6,6}}],Polygon[{xx[1],xx[2],xx[3],xx[4]}],Polygon[{xx[1],xx[2],rotationmat[α,xx[2],xx[1],xx[2]+{x,-0,0}],rotationmat[α,xx[2],xx[1],xx[1]+{x,0,0}]}],
Polygon[{xx[4],xx[1],rotationmat[α,xx[1],xx[4],xx[1]+{0,x,0}],rotationmat[α,xx[1],xx[4],xx[4]+{-0,x,0}]}],
Polygon[{xx[3],xx[4],rotationmat[α,xx[4],xx[3],xx[4]+{-x,0,0}],rotationmat[α,xx[4],xx[3],xx[3]+{-x,-0,0}]}],Polygon[{xx[2],xx[3],rotationmat[α,xx[3],xx[2],xx[3]+{-0,-x,0}],rotationmat[α,xx[3],xx[2],xx[2]+{0,-x,0}]}]},Background->Black,AspectRatio->Automatic]];
Manipulate[Which[nss==1,zukei2Db[x2],nss==2,zukei3Db[x2,βb2,0],nss==3,α02=ArcCos[0/x2]//N;zukei3Db[x2,α02,1]],
{{nss,1,"Cup, Making of cup 3D, Making of cup 2D"},Range[1,3],ControlType->Setter},
{{βb2,0,"β"},0,Pi/2+0.015,0.001,Appearance->"Labeled",ImageSize->Tiny},
{{x2,2.73,"x"},0.1,5.99,0.01,Appearance->"Labeled",ImageSize->Tiny},
SaveDefinitions->True,ControlPlacement->Left]

この問題の答は簡単です。容積V=x(((12-2x))^2)のグラフは0<x<6で次のようになるので、最大値はx=2のとき、V=128となる。

The Graph of y=x(((12-2x))^2)  for 0 < x < 6, and the max value is 128.

普通は、数学の問題としてはここで終わりますが、私達は、ここからスタートします。

この問題から出発して新しい問題を考えることができますか？

Usually we finish this problem, and move to another problem, but we are to make a new problem based on this.

Can you make a new problem out of this problem?

新しい問題を考えるというと、難しいように思われるかもしれません。でも、生徒達に正方形の紙で、コップを作るように言うとすぐに作ります。

It may seem to be quite difficult to think up a new problem, but young students find it easy to make a cup of various types.

全く自由に作ってもらうと次のようになります。

If they are free to make any kind of cup, they will present many kinds of cups.

ある生徒が次のようなタイプのコップを考えました。xとyに注意してください。

A student proposed the following cup.

Clear[y];zukei2D[x_,y_]:=Block[{xx2},xx2[1]={6-x,6-x};xx2[2]={6-x,-6+x};xx2[3]={-6+x,-6+x};xx2[4]={-6+x,6-x};pts0={{6-x,6},{5-x,6-x/2},{6-x,6-x}};pts1={{6-x,6},{6-x+y/2,6-x/4},{6-x+y,6}};pts2={{6-x,6},{6-x-1,0},{6-x,-6}};
Graphics[{{Gray,Polygon[{{6,6},{5.99,6},{5.99,6}}]},Polygon[{{-6,6},{-5.99,6},{-5.99,6}}],Polygon[{{6,-6},{5.99,-6},{5.99,-6}}],Polygon[{{-6,-6},{-5.99,-6},{-5.99,-6}}],
Polygon[{{6,6},{5.99,6},{5.99,6}}],Polygon[{{-6,6},{-5.99,6},{-5.99,6}}],Polygon[{{6,-6},{5.99,-6},{5.99,-6}}],Polygon[{{-6,-6},{-5.99,-6},{-5.99,-6}}],{RGBColor[0.6,0.9,0.9],Polygon[{xx2[1],xx2[2],xx2[3],xx2[4]}]},{RGBColor[0.6,0.9,0.9],Polygon[{xx2[1],xx2[2],xx2[2]+{x,-y},xx2[1]+{x,y}}]},
{RGBColor[0.6,0.9,0.9],Polygon[{xx2[4],xx2[1],xx2[1]+{y,x},xx2[4]+{-y,x}}]},
{RGBColor[0.6,0.9,0.9],Polygon[{xx2[3],xx2[4],xx2[4]+{-x,y},xx2[3]+{-x,-y}}]},{RGBColor[0.6,0.9,0.9],Polygon[{xx2[2],xx2[3],xx2[3]+{-y,-x},xx2[2]+{y,-x}}]},{GrayLevel[0.5],BSplineCurve[pts2]},{GrayLevel[0.5],BSplineCurve[pts0]},{GrayLevel[0.5],BSplineCurve[pts1]},{Black,Text["xcm",{5-x,6-x/2}]},If[y>0.1,{Black,Text["ycm",{6-x+y/3,6-x/4}]},{RGBColor[0.6,0.9,0.9],Text["ycm",{0,0}]}],{Black,Text["12cm",{6-x-1,0}]}},Background->Black,AspectRatio->Automatic]];rotationmat[dg_,point1_,point2_,rotatepoint_]:=Block[{degr=dg,p1=point1,p2=point2,p3=rotatepoint},(RotationMatrix[dg,p1-p2].(p3-p1))+p1];zukei3D[x_,y_,α_,gr_]:=Block[{xx,h},xx[1]={6-x,6-x,0};xx[2]={6-x,-6+x,0};xx[3]={-6+x,-6+x,0};xx[4]={-6+x,6-x,0};h=(8/6) ((((x-y))^2)+(x-y) *x-18(x-y)-18 x+(x^2)+108)(Sqrt[(x^2)-(((y))^2)])//N;
Graphics3D[{If[gr==1,{White,Text[h,{3,3,6}]},{Black,Text[h,{3,3,6}]}],Polygon[{{6,6,0},{5.99,6,0},{5.99,6,0}}],Polygon[{{-6,6,0},{-5.99,6,0},{-5.99,6,0}}],Polygon[{{6,-6,0},{5.99,-6,0},{5.99,-6,0}}],Polygon[{{-6,-6,0},{-5.99,-6,0},{-5.99,-6,0}}],
Polygon[{{6,6,6},{5.99,6,6},{5.99,6,6}}],Polygon[{{-6,6,6},{-5.99,6,6},{-5.99,6,6}}],Polygon[{{6,-6,6},{5.99,-6,6},{5.99,-6,6}}],Polygon[{{-6,-6,6},{-5.99,-6,6},{-5.99,-6,6}}],Polygon[{xx[1],xx[2],xx[3],xx[4]}],Polygon[{xx[1],xx[2],rotationmat[α,xx[2],xx[1],xx[2]+{x,-y,0}],rotationmat[α,xx[2],xx[1],xx[1]+{x,y,0}]}],
Polygon[{xx[4],xx[1],rotationmat[α,xx[1],xx[4],xx[1]+{y,x,0}],rotationmat[α,xx[1],xx[4],xx[4]+{-y,x,0}]}],
Polygon[{xx[3],xx[4],rotationmat[α,xx[4],xx[3],xx[4]+{-x,y,0}],rotationmat[α,xx[4],xx[3],xx[3]+{-x,-y,0}]}],Polygon[{xx[2],xx[3],rotationmat[α,xx[3],xx[2],xx[3]+{-y,-x,0}],rotationmat[α,xx[3],xx[2],xx[2]+{y,-x,0}]}]},Background->Black,AspectRatio->Automatic]];
Manipulate[Which[ns==1,zukei2D[x,y],ns==2,zukei3D[x,y,βb,0],ns==3,α0=ArcCos[y/x]//N;zukei3D[x,y,α0,1]],
{{ns,1,"Cup, Making of cup 3D, Making of cup 2D"},Range[1,3],ControlType->Setter},
{{βb,0,"β"},0,Pi/2+0.015,0.001,Appearance->"Labeled",ImageSize->Tiny},
{{x,1.73,"x"},0.1,5.99,0.01,Appearance->"Labeled",ImageSize->Tiny},
{{y,1.43,"y"},0,x,0.01,Appearance->"Labeled",ImageSize->Tiny},
SaveDefinitions->True,ControlPlacement->Left]

容積( volume) V=(4/3)((x^2)+xy-18x-18y+(y^2)+108)(Sqrt[(y^2)-(((y-x))^2)])からMathematicaで最大を計算してみます。

Next, we calculate the maximum of volume.

手順は、Vをxとyで微分して、それらを0とおきます。
xとyに関する連立方程式で文字xを消去すると、次のようになります。

49 (y^6)-1400 (y^5)+16284 (y^4)-98928 (y^3)+331776 (y^2)-583200 y+419904=0

y=6は容器が作れない。7 (y^2)-54 y+108=0は虚数解(imaginary numbers)なので、

(y-6) () (7 (y^3)-104 (y^2)+468 y-648)=0

7 (y^3)-104 (y^2)+468 y-648=0の数値解はy=2.78616,4.24616, 7.82482で、

y=2.786159961512071`,x=1.306170972327122` のとき、V= 149.333(cm^3)

Mathematicaの計算を紹介します。

Next, we show the calculation by Mathematica.

Clear[x,y];D[(4/3)((x^2)+x y-18x-18y+(y^2)+108)(Sqrt[(y^2)-(((y-x))^2)]),x]//Simplify

Clear[x,y];D[(4/3)((x^2)+x y-18x-18y+(y^2)+108)(Sqrt[(y^2)-(((y-x))^2)]),y]//Simplify

Solve[108+(-54+x) y+5 (y^2)==0,x]

x=((-108+54 y-5 (y^2))/y);3 (x^3)-3 (x^2) (12+y)-2 x (-54-18 y+(y^2))-y (108-18 y+(y^2))//Simplify

Factor[419904-583200 y+331776 (y^2)-98928 (y^3)+16284 (y^4)-1400 (y^5)+49 (y^6)]

(-6+y) (108-54 y+7 (y^2)) (-648+468 y-104 (y^2)+7 (y^3))

Solve[108-54 y+7 (y^2)==0,y]

NSolve[-648+468 y-104 (y^2)+7 (y^3)==0,y]

Clear[x,y];y=2.7861599615120665`;x=((-108+54 y-5 (y^2))/y);(4/3)((x^2)+x y-18x-18y+(y^2)+108)(Sqrt[(y^2)-(((y-x))^2)])

x=((-108+54 y-5 (y^2))/y)

Clear[x,y];y=4.246161014548971`;x=((-108+54 y-5 (y^2))/y);(4/3)((x^2)+x y-18x-18y+(y^2)+108)(Sqrt[(y^2)-(((y-x))^2)])

Clear[x,y];y=7.824821881081822`;x=((-108+54 y-5 (y^2))/y);(4/3)((x^2)+x y-18x-18y+(y^2)+108)(Sqrt[(y^2)-(((y-x))^2)])

When x=1.306170972327122` and y=2.786159961512071`, 
V= 149.333(cm^3)

以上の内容はこの雑誌で掲載されました。

Our result on Origami Cup was published by this journal

この研究は良い結果を出した例ですが、良い結果を出した理由を考えてみます。

This is a successful example of our research. Is there any reason why it was successful?

元々の問題は正方形の紙の四隅を正方形に切り取って、残った紙を折り曲げて、直方体を作り、容積をできる限り大きくしました。

We cut out four squares from the corners of a square paper to make a cup of cuboid, and make it as large as possible.

元々の問題は正方形の紙の四隅を正方形に切り取って、残った紙を折り曲げて、直方体を作り、容積をできる限り大きくしました。

In the original problem, we cut out four squares from the corners of a square paper to make a cup of cuboid, and make it as large as possible.

問題を正確に書いてみて、どこを変更できるかを考えさせると、たいていの生徒はいろんな問題の変形を思いつきます。

If a problem is expressed in a precise sentence, then it becomes quite easy for most of the students to create new problem based on the original one.

次に少し変わったコップを紹介します。ある生徒に次のようなコップを見せたら、このようにしました。

Here, I would like to introduce a very unique cup. When I show the following cup to one of my students, he ...

Manipulate[
α=Min[-1+1*0.1t,0];
θ=Min[0+1.11*((Ceiling[(t-10)/10.1]*(t-10))/10),1.11];
β=Max[2.41+-2.41*((Ceiling[(t-20)/20.1]*(t-20))/10),0];
γ=Min[0+3.2*((Ceiling[(t-30)/30.1]*(t-30))/10),3.2];
If[t<10.5,Graphics3D[{Polygon[{{6,6,6},{6.1,6.1,6.1},{6.2,6.2,6.2}}],Polygon[-{{6,6,6},{6.1,6.1,6.1},{6.2,6.2,6.2}}],Polygon[{{-6,6,6},{-6.1,6.1,6.1},{-6.2,6.2,6.2}}],Polygon[{{6,-6,6},{6.1,-6.1,6.1},{6.2,-6.2,6.2}}],
Polygon[{{x,0,0}+p,{x,0,0}+r,rotationmat[α,{x,0,0}+p,{x,0,0}+r,{x,0,0}+r+{y-x,y-x,(Sqrt[y^2-(y-x)^2])}],
rotationmat[α,{x,0,0}+p,{x,0,0}+r,{x,0,0}+p+{y-x,-y+x,(Sqrt[y^2-(y-x)^2])}]}],
Polygon[{{x,0,0}+r,{-x,0,0}+u,rotationmat[α,{x,0,0}+r,{-x,0,0}+u,{-x,0,0}+u+{-y+x,y-x,(Sqrt[y^2-(y-x)^2])}],rotationmat[α,{x,0,0}+r,{-x,0,0}+u,{x,0,0}+r+{y-x,y-x,(Sqrt[y^2-(y-x)^2])}]}],
Polygon[{{-x,0,0}+u,{-x,0,0}+s,rotationmat[α,{-x,0,0}+u,{-x,0,0}+s,{-x,0,0}+s+{-y+x,-y+x,(Sqrt[y^2-(y-x)^2])}],rotationmat[α,{-x,0,0}+u,{-x,0,0}+s,{-x,0,0}+u+{-y+x,y-x,(Sqrt[y^2-(y-x)^2])}]}],
Polygon[{{x,0,0}+p,{x,0,0}+r,{-x,0,0}+u,{-x,0,0}+s}],
Polygon[{{x,0,0}+p,{-x,0,0}+s,rotationmat[-α,{x,0,0}+p,{-x,0,0}+s,{-x,0,0}+s+{-y+x,-y+x,(Sqrt[y^2-(y-x)^2])}],rotationmat[-α,{x,0,0}+p,{-x,0,0}+s,{x,0,0}+p+{y-x,-y+x,(Sqrt[y^2-(y-x)^2])}]}]},AspectRatio->1],Graphics3D[{Polygon[{{6,6,6},{6.1,6.1,6.1},{6.2,6.2,6.2}}],Polygon[-{{6,6,6},{6.1,6.1,6.1},{6.2,6.2,6.2}}],Polygon[{{-6,6,6},{-6.1,6.1,6.1},{-6.2,6.2,6.2}}],Polygon[{{6,-6,6},{6.1,-6.1,6.1},{6.2,-6.2,6.2}}],
Polygon[{{x,0,0}+p,{x,0,0}+r,rotationmat[α,{x,0,0}+p,{x,0,0}+r,{x,0,0}+r+{y-x,y-x,(Sqrt[y^2-(y-x)^2])}],rotationmat[α,{x,0,0}+p,{x,0,0}+r,{x,0,0}+p+{y-x,-y+x,(Sqrt[y^2-(y-x)^2])}]}],
Polygon[{{x,0,0}+p,{x,0,0}+r,rotationmat[θ,{x,0,0}+p,{x,0,0}+r,({x,0,0}+p+{-x,0,0}+u)/2]}],
Polygon[{{x,0,0}+r,{-x,0,0}+u,rotationmat[θ,{x,0,0}+r,{-x,0,0}+u,({-x,0,0}+u+{x,0,0}+p)/2]}],
Polygon[{{-x,0,0}+u,{-x,0,0}+s,rotationmat[θ,{-x,0,0}+u,{-x,0,0}+s,({-x,0,0}+s+{x,0,0}+r)/2]}],
Polygon[{{x,0,0}+p,{-x,0,0}+s,rotationmat[θ,{-x,0,0}+s,{x,0,0}+p,({-x,0,0}+s+{x,0,0}+r)/2]}],
Polygon[{{x,0,0}+r,{-x,0,0}+u,rotationmat[α,{x,0,0}+r,{-x,0,0}+u,{-x,0,0}+u+{-y+x,y-x,(Sqrt[y^2-(y-x)^2])}],rotationmat[α,{x,0,0}+r,{-x,0,0}+u,{x,0,0}+r+{y-x,y-x,(Sqrt[y^2-(y-x)^2])}]}],
Polygon[{{-x,0,0}+u,{-x,0,0}+s,rotationmat[α,{-x,0,0}+u,{-x,0,0}+s,{-x,0,0}+s+{-y+x,-y+x,(Sqrt[y^2-(y-x)^2])}],rotationmat[α,{-x,0,0}+u,{-x,0,0}+s,{-x,0,0}+u+{-y+x,y-x,(Sqrt[y^2-(y-x)^2])}]}],
Polygon[{{x,0,0}+p,{-x,0,0}+s,rotationmat[-α,{x,0,0}+p,{-x,0,0}+s,{-x,0,0}+s+{-y+x,-y+x,(Sqrt[y^2-(y-x)^2])}],rotationmat[-α,{x,0,0}+p,{-x,0,0}+s,{x,0,0}+p+{y-x,-y+x,(Sqrt[y^2-(y-x)^2])}]}]},AspectRatio->1]],
{{t,10.3,"t"},10.3,40,0.1,Appearance->"Labeled",ImageSize->Tiny},
SaveDefinitions->True,ControlPlacement->Left]

彼はこのようなことをした後で、用事を思い出して去って行きました。そのあとを別の生徒に頼みました。

After he did this kind of thing, he went away. I asked another student to do something about this.

Manipulate[
α1=Min[-1+1*0.1t1,0];
θ1=Min[0+1.11*((Ceiling[(t1-10)/10.1]*(t1-10))/10),1.11];
β1=Max[2.41+-2.41*((Ceiling[(t1-20)/20.1]*(t1-20))/10),0];
γ1=Min[0+3.2*((Ceiling[(t1-30)/30.1]*(t1-30))/10),3.2];
Graphics3D[{Polygon[{{6,6,6},{6.1,6.1,6.1},{6.2,6.2,6.2}}],Polygon[-{{6,6,6},{6.1,6.1,6.1},{6.2,6.2,6.2}}],Polygon[{{-6,6,6},{-6.1,6.1,6.1},{-6.2,6.2,6.2}}],Polygon[{{6,-6,6},{6.1,-6.1,6.1},{6.2,-6.2,6.2}}],Polygon[{rotationmat[β1,{2x,-x,0}+r,{0,x,0}+r,{x,0,0}],rotationmat[β1,{2x,-x,0}+r,{0,x,0}+r,{0,x,0}],{x,0,0}+r}],
Polygon[{rotationmat[-β1,{2x,x,0}+p,{0,-x,0}+p,{x,0,0}],rotationmat[-β1,{2x,x,0}+p,{0,-x,0}+p,{0,-x,0}],{x,0,0}+p}],
Polygon[{rotationmat[-β1,{0,-x,0}+s,{-2x,x,0}+s,{0,-x,0}],rotationmat[-β1,{0,-x,0}+s,{-2x,x,0}+s,{-x,0,0}],{-x,0,0}+s}],
Polygon[{rotationmat[β1,{0,x,0}+u,{-2x,-x,0}+u,{0,x,0}],rotationmat[β1,{0,x,0}+u,{-2x,-x,0}+u,{-x,0,0}],{-x,0,0}+u}],
Polygon[{rotationmat[β1,{2x,-x,0}+r,{0,x,0}+r,{x,0,0}],rotationmat[β1,{2x,-x,0}+r,{0,x,0}+r,{0,x,0}],rotationmat[-γ1,rotationmat[β1,{2x,-x,0}+r,{0,x,0}+r,{x,0,0}],rotationmat[β1,{2x,-x,0}+r,{0,x,0}+r,{0,x,0}],(rotationmat[β1,{2x,-x,0}+r,{0,x,0}+r,{x,0,0}]+rotationmat[β1,{2x,-x,0}+r,{0,x,0}+r,{0,x,0}])/2+{x,x,0}/2]}],
Polygon[{rotationmat[-β1,{2x,x,0}+p,{0,-x,0}+p,{x,0,0}],rotationmat[-β1,{2x,x,0}+p,{0,-x,0}+p,{0,-x,0}],rotationmat[γ1,rotationmat[-β1,{2x,x,0}+p,{0,-x,0}+p,{x,0,0}],rotationmat[-β1,{2x,x,0}+p,{0,-x,0}+p,{0,-x,0}],(rotationmat[-β1,{2x,x,0}+p,{0,-x,0}+p,{x,0,0}]+rotationmat[-β1,{2x,x,0}+p,{0,-x,0}+p,{0,-x,0}])/2+{x,-x,0}/2]}],
Polygon[{rotationmat[-β1,{0,-x,0}+s,{-2x,x,0}+s,{0,-x,0}],rotationmat[-β1,{0,-x,0}+s,{-2x,x,0}+s,{-x,0,0}],rotationmat[γ1,rotationmat[-β1,{0,-x,0}+s,{-2x,x,0}+s,{0,-x,0}],rotationmat[-β1,{0,-x,0}+s,{-2x,x,0}+s,{-x,0,0}],(rotationmat[-β1,{0,-x,0}+s,{-2x,x,0}+s,{0,-x,0}]+rotationmat[-β1,{0,-x,0}+s,{-2x,x,0}+s,{-x,0,0}])/2+{-x,-x,0}/2]}],
Polygon[{rotationmat[β1,{0,x,0}+u,{-2x,-x,0}+u,{0,x,0}],rotationmat[β1,{0,x,0}+u,{-2x,-x,0}+u,{-x,0,0}],rotationmat[-γ1,rotationmat[β1,{0,x,0}+u,{-2x,-x,0}+u,{0,x,0}],rotationmat[β1,{0,x,0}+u,{-2x,-x,0}+u,{-x,0,0}],(rotationmat[β1,{0,x,0}+u,{-2x,-x,0}+u,{0,x,0}]+rotationmat[β1,{0,x,0}+u,{-2x,-x,0}+u,{-x,0,0}])/2+{-x,x,0}/2]}],
Polygon[{{x,0,0}+p,{x,0,0}+r,rotationmat[α1,{x,0,0}+p,{x,0,0}+r,{x,0,0}+r+{y-x,y-x,(Sqrt[y^2-(y-x)^2])}],rotationmat[α1,{x,0,0}+p,{x,0,0}+r,{x,0,0}+p+{y-x,-y+x,(Sqrt[y^2-(y-x)^2])}]}],Polygon[{{x,0,0}+p,{x,0,0}+r,rotationmat[θ1,{x,0,0}+p,{x,0,0}+r,({x,0,0}+p+{-x,0,0}+u)/2]}],
Polygon[{{x,0,0}+r,{-x,0,0}+u,rotationmat[α1,{x,0,0}+r,{-x,0,0}+u,{-x,0,0}+u+{-y+x,y-x,(Sqrt[y^2-(y-x)^2])}],rotationmat[α1,{x,0,0}+r,{-x,0,0}+u,{x,0,0}+r+{y-x,y-x,(Sqrt[y^2-(y-x)^2])}]}],
Polygon[{{x,0,0}+r,{-x,0,0}+u,rotationmat[θ1,{x,0,0}+r,{-x,0,0}+u,({-x,0,0}+u+{x,0,0}+p)/2]}],Polygon[{{-x,0,0}+u,{-x,0,0}+s,rotationmat[α1,{-x,0,0}+u,{-x,0,0}+s,{-x,0,0}+s+{-y+x,-y+x,(Sqrt[y^2-(y-x)^2])}],rotationmat[α1,{-x,0,0}+u,{-x,0,0}+s,{-x,0,0}+u+{-y+x,y-x,(Sqrt[y^2-(y-x)^2])}]}],
Polygon[{{-x,0,0}+u,{-x,0,0}+s,rotationmat[θ1,{-x,0,0}+u,{-x,0,0}+s,({-x,0,0}+s+{x,0,0}+r)/2]}],Polygon[{{x,0,0}+p,{-x,0,0}+s,rotationmat[-α1,{x,0,0}+p,{-x,0,0}+s,{-x,0,0}+s+{-y+x,-y+x,(Sqrt[y^2-(y-x)^2])}],rotationmat[-α1,{x,0,0}+p,{-x,0,0}+s,{x,0,0}+p+{y-x,-y+x,(Sqrt[y^2-(y-x)^2])}]}],Polygon[{{x,0,0}+p,{-x,0,0}+s,rotationmat[θ1,{-x,0,0}+s,{x,0,0}+p,({-x,0,0}+s+{x,0,0}+r)/2]}]},AspectRatio->1],
{{t1,0,"t"},0,40,0.1,Appearance->"Labeled",ImageSize->Tiny},
SaveDefinitions->True,ControlPlacement->Left]

この図形で私達のグループはGoogle Science Fairのアジア代表になり、テレビ出演もしました。

With this my students represent Asia at Google Science Fair. They also showed this figure on TV.

例　　カードゲームの確率（ロシアンルーレット）
赤のカードが2枚と白のカードが4枚、合計6枚のカードがある。二人のプレイヤーA,Bが交互にカードを１枚引くが、引いたカードは元に戻さない。先に赤のカードを引いたプレイヤーが負ける。先手のプレイヤーが負ける確率を求る。

(元々はRussian Rouletteの問題だった．)

There are two red cards and four white card. Two players A and B take turns to pick a card.The first player to draw out a red card loses the game.Calculate the probability that the first player will lose the game.(Originally we studied this as a Russian roulette game)

先手が負ける確率は(the first player’s losing ) 
(2/6)+(4/6)×(3/5)×(2/4)+(4/6)×(3/5)×(2/4)×(1/3)×(2/2)=(3/5).

赤のカードが2枚と白のカードが4枚、合計6枚のカードがある。二人のプレイヤー

There are two red cards and four white card. Two players

パスカルの三角形に似た、分数からできた三角形
A triangle of fractions that looks like the Pascal’s triangle.

Clear[t,p,v];triangle[t_,p_,v_]:=Block[{pro},pro[p2_,n2_,m2_,v2_]:=Block[{t2,u},t2=IntegerPart[((1+p2-m2+n2-v2)/p2)];u=((∑_(z=0)^(t2-1))Binomial[n2-v2-p2*z,m2-1]);FractionBox[u,Binomial[n2,m2]]//DisplayForm];Table[Table[pro[p,n,m,v],{m,1,n}],{n,1,t}]];Manipulate[If[v>p,v=1];Graphics[Text@Style[Column[Map[Row[#,Spacer[N@Log[2,t]]]&,triangle[t,p,
v]],Center,Spacings->1/2],18],ImageSize->{650,600}],
{{t,3,"number of rows n"},2,12,1,Appearance->"Labeled"},
{{p,2,"number of players p"},2,10,1,Appearance->"Labeled"},
{{v,1,"player v"},1,p,1,Appearance->"Labeled"},
SaveDefinitions->True,ControlPlacement->Left,TrackedSymbols->{t,p,v},AutorunSequencing->{1}]

分数が作る三角形のMathematicaのコードを見ていただきます。

I am going to show Mathematica codes for this triangle of fractions.

triangle[t_,p_,v_]:=Block[{pro},pro[p2_,n2_,m2_,v2_]:=Block[{t2,u},t2=IntegerPart[((1+p2-m2+n2-v2)/p2)];u=((∑_(z=0)^(t2-1))Binomial[n2-v2-p2*z,m2-1]);FractionBox[u,Binomial[n2,m2]]//DisplayForm];Table[Table[pro[p,n,m,v],{m,1,n}],{n,1,t}]];Manipulate[If[v>p,v=1];Graphics[Text@Style[Column[Map[Row[#,Spacer[N@Log[2,t]]]&,triangle[t,p,
v]],Center,Spacings->1/2],18],ImageSize->{650,600}],
{{t,3,"number of rows n"},2,12,1,Appearance->"Labeled"},
{{p,2,"number of players p"},2,10,1,Appearance->"Labeled"},
{{v,1,"player v"},1,p,1,Appearance->"Labeled"},
SaveDefinitions->True,TrackedSymbols->{t,p,v},AutorunSequencing->{1}]

This result was published by Mathematical Gazette (England)

例　　カードゲームの確率（ロシアンルーレット）
赤のカードが2枚と白のカードが4枚、合計6枚のカードがある。二人のプレイヤーA,Bが交互にカードを１枚引くが、引いたカードは元に戻さない。先に赤のカードを引いたプレイヤーが負ける。

There are two red cards and four white card. Two players A and B take turns to pick a card.The first player to draw out a red card ...

Clear[triangle,ft];triangle[t_,ss1_,ss2_,gg1_,gg2_]:=Block[{ft},ft[n_,m_,s1_,s2_,g1_,g2_]:=FractionBox[((∑_(h=1)^s1)(∑_(k=Ceiling[(g1-h+s1)/s1])^(Min[Floor[(n-h+s1+s2)/(s1+s2)],Floor[(n-m+g1-h+s1)/s1],Floor[(n-m+g1+g2-1-h+s1+s2)/(s1+s2)]]))(∑_(u=Max[0,m-n-g1+(k-1)(s1+s2)+h])^(Min[s2(k-1),g2-1]))(Binomial[s1(k-1)+h-1,g1-1]*Binomial[s2(k-1),u]*Binomial[n-(k-1)(s1+s2)-h,m-g1-u])),Binomial[n,m]]//DisplayForm;
Table[Table[ft[n,m,ss1,ss2,gg1,gg2],{m,1,n}],{n,1,t}]];Manipulate[Graphics[Text@Style[Column[Map[Row[#,Spacer[N@Log[2,8t]]]&,triangle[t,s1,s2,g1,g2]],Center,Spacings->1/2.3],18],ImageSize->{650,620}],
{{t,11,"number of rows n"},2,20,1,Appearance->"Labeled"},
{{s1,2,"s1"},1,6,1,Appearance->"Labeled"},
{{s2,4,"s2"},1,6,1,Appearance->"Labeled"},
{{g1,4,"g1"},1,6,1,Appearance->"Labeled"},
{{g2,2,"g2"},1,6,1,Appearance->"Labeled"},
SaveDefinitions->True,ControlPlacement->Left,TrackedSymbols->{t,s1,s2,g1,g2},AutorunSequencing->{1}]

JCDCG^3  in Tokyo

紙を切って作るコップの問題と、ロシアンルーレットから始まったカードゲームのことを話しましたが、そこで使った方法を振り返ってみます。

I have talked about a cup made from a paper, and cards game originated from Russian Roulette. Let’s review the method used in these research.

1.適当な問題を紹介　(Introduction of proper problems for students)
2. それを解くのではなく、変形する (change the problems instead of solving them)
3. 変形した問題を数学ソフトを使って調べる (study with Mathematica)
4. 数学ソフトの計算結果を分析　(Study the result of calculation)
5. 面白い結果が出たらそれを証明 (If you find some formula, prove them)
6. 学会発表や専門誌への投稿、コンテストへの参加 (Talk at conferences, submit papers and participate in contests)

注意　　問題を書いてみて、どの部分を変更できるかを考えたら、変形は簡単です。証明に関しては、高校生でもできる場合と、専門家の助けが必要な場合もあります。

チョコレート問題グレーの部分は甘く，黒の部分は苦いチョコレートがある．2人のプレイヤーが交互に線に沿って切り、食べる。
最後に苦い部分を残されたら負ける．先手と後手のどちらが有利か．

The chocolate consists of sweet parts and a bitter part.Two players take turns, and cut the chocolate along the straight line into two areas, and eat the area without bitter part.

DynamicModule[{cDraw,cutC,tCount,xs,ys,zs,mp,end},
Manipulate[
Row[{
Column[{

Text[Style[StringForm["Coordinates (x, y, z) = (``, ``, ``)",xs,ys,zs],FontSize->24]],
If[{xs,ys,zs}=={0,0,0},end=1;
Text[Style[StringForm[" win ",If[Mod[tCount,2]==1,1,2]],FontSize->24]],
Text[Style[StringForm["Player``",If[end==1,end=0;tCount=0];If[Mod[tCount,2]==0,1,2]],FontSize->24]]
],
Dynamic@EventHandler[cDraw[xs,ys,zs],
"MouseDown":>(mp=MousePosition["Graphics"];{xs,ys,zs}=cutC[mp,xs,ys,zs])
]
},Center]
}],
{{xs,4,"x"},0,4,1},
{{ys,5,"y"},0,5,1},
{{zs,5,"z"},0,5,1},
ControlPlacement->Left],
Initialization:>(
tCount=0;xs=4;ys=5;zs=5;end=0;
cDraw[x_,y_,z_]:=Graphics[{EdgeForm[Thick],
{RGBColor[0.2,0.4,0.8],Polygon[{{0,0},{0,1},{1,1},{1,0}}]},
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{0,r},{0,r+1},{1,r+1},{1,r}}]},{r,1,y}],
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{-p,q},{-p,q+1},{-p+1,q+1},{-p+1,q}}]},{p,1,x},{q,0,y}],
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{+p,q},{+p,q+1},{+p+1,q+1},{+p+1,q}}]},{p,1,z},{q,0,y}]
},Background->Gray,ImageSize->{540,315},PlotRange->{{-5.5,6.5},{-0.5,6.5}}];
cutC[p_,x_,y_,z_]:=Block[{m1,m2,n,cx,cy,cz},
m1=p[[1]];m2=p[[2]];n=0.2;cx=0;cy=0;cz=0;
If[-x-1<m1<z+2&&-1<m2<y+2,
Do[If[-v-n<m1-0<-v+n,cx=x-v],{v,0,x}];
Do[If[+v-n<m1-1<+v+n,cz=z-v],{v,0,z}];
Do[If[+v-n<m2-1<+v+n,cy=y-v],{v,0,y}];
];
If[cx+cy+cz≠0&&(cx+cy==0||cy+cz==0||cz+cx==0),
tCount++;return={x-cx,y-cy,z-cz},
return={x,y,z}
];
return
];
)
]

このチョコレートも数学的には同じ．

This chocolate is mathematically the same.

DynamicModule[{cDraw,cutC,tCount,xss,yss,zss,mp,end},
Manipulate[
Row[{
Column[{

Text[Style[StringForm["Coordinates (\!\(\*
StyleBox[\"x\",\nFontFamily->\"Times New Roman Italic\",\nFontWeight->\"Plain\"]\)\!\(\*
StyleBox[\",\",\nFontFamily->\"Times New Roman Italic\",\nFontWeight->\"Plain\"]\)\!\(\*
StyleBox[\" \",\nFontFamily->\"Times New Roman Italic\",\nFontWeight->\"Plain\"]\)\!\(\*
StyleBox[\"y\",\nFontFamily->\"Times New Roman Italic\",\nFontWeight->\"Plain\"]\)\!\(\*
StyleBox[\",\",\nFontFamily->\"Times New Roman Italic\",\nFontWeight->\"Plain\"]\)\!\(\*
StyleBox[\" \",\nFontFamily->\"Times New Roman Italic\",\nFontWeight->\"Plain\"]\)\!\(\*
StyleBox[\"z\",\nFontFamily->\"Times New Roman Italic\",\nFontWeight->\"Plain\"]\)) = (``, ``, ``)",xss,yss,zss],FontSize->24]],
If[{xss,yss,zss}=={0,0,0},end=1;
Text[Style[StringForm[" win ",If[Mod[tCount,2]==1,1,2]],FontSize->24]],
Text[Style[StringForm["Player``",If[end==1,end=0;tCount=0];If[Mod[tCount,2]==0,1,2]],FontSize->24]]
],
Dynamic@EventHandler[ccDraw[xss,yss,zss],
"MouseDown":>(mp=MousePosition["Graphics"];{xss,yss,zss}=cutC[mp,xss,yss,zss])
]
},Center]
}],
{{xss,4,"x"},0,4,1},
{{yss,5,"y"},0,5,1},
{{zss,5,"z"},0,5,1},
ControlPlacement->Left],
Initialization:>(
tCount=0;xss=4;yss=5;zss=5;end=0;
ccDraw[x_,y_,z_]:=Graphics[{EdgeForm[Thick],
{RGBColor[0.2,0.4,0.8],Polygon[{{0,0},{0,1},{1,1},{1,0}}]},
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{0,r},{0,r+1},{1,r+1},{1,r}}]},{r,1,y}],
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{-p,q},{-p,q+1},{-p+1,q+1},{-p+1,q}}]},{p,1,x},{q,0,0}],
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{+p,q},{+p,q+1},{+p+1,q+1},{+p+1,q}}]},{p,1,z},{q,0,0}]
},Background->Gray,ImageSize->{540,315},PlotRange->{{-5.5,6.5},{-0.5,6.5}}];
cutC[p_,x_,y_,z_]:=Block[{m1,m2,n,cx,cy,cz},
m1=p[[1]];m2=p[[2]];n=0.2;cx=0;cy=0;cz=0;
If[-x-1<m1<z+2&&-1<m2<y+2,
Do[If[-v-n<m1-0<-v+n,cx=x-v],{v,0,x}];
Do[If[+v-n<m1-1<+v+n,cz=z-v],{v,0,z}];
Do[If[+v-n<m2-1<+v+n,cy=y-v],{v,0,y}];
];
If[cx+cy+cz≠0&&(cx+cy==0||cy+cz==0||cz+cx==0),
tCount++;return={x-cx,y-cy,z-cz},
return={x,y,z}
];
return
];
)
]

このチョコレートも数学的には同じだが、これは普通の3つの山の石とりゲームである。

This chocolate is mathematically the same, but this is the traditional game of NIM with three piles.

DDDraw[x_,y_,z_]:=Graphics[{EdgeForm[Thick],

Table[{RGBColor[0.6,0.4,0.2],Polygon[{{-2,r},{-2,r+1},{-1,r+1},{-1,r}}]},{r,1,x}],
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{2,r},{2,r+1},{3,r+1},{3,r}}]},{r,1,z}],
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{0,r},{0,r+1},{1,r+1},{1,r}}]},{r,1,y}],{RGBColor[0.2,0.4,0.8],Polygon[{{-2,0},{-2,1},{3,1},{3,0}}]}
},Background->Gray,ImageSize->{540,515},PlotRange->{{-5.5,6.5},{-0.5,6.5}}];Manipulate[DDDraw[x,y,z],{{x,4,"x"},0,4,1},{{y,5,"y"},0,5,1},{{z,5,"z"},0,5,1}]

このチョコレートも数学的に異なる。それは、縦に切ると、高さが同時に変わる。

Mathematical structure is different, since a vertical break can reduce the number of horizontal breaks.

DynamicModule[{cDraw,cutC,tCount,xs,ys,zs,mp,end},
Manipulate[
Row[{
Column[{
ys=Min[ys,zs];
Text[Style[StringForm["coordinates (x, y, z) = (``, ``, ``)",xs,ys,zs],FontSize->36]],
Dynamic@EventHandler[cDraw[xs,ys,zs],
"MouseDown":>(mp=MousePosition["Graphics"];{xs,ys,zs}=cutC[mp,xs,ys,zs])
]
},Center]
}],
{{xs,7,"x"},0,7,1},
{{ys,7,"y"},0,7,1},
{{zs,7,"z"},0,7,1},
ControlPlacement->Left],
Initialization:>(
tCount=0;xs=5;ys=5;zs=5;end=0;
cDraw[x_,y_,z_]:=Graphics[{EdgeForm[Thick],
{RGBColor[0.2,0.4,0.8],Polygon[{{0,0},{0,1},{1,1},{1,0}}]},
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{-p,0},{-p,1},{-p+1,1},{-p+1,0}}]},{p,1,x}],
Table[{RGBColor[0.6,0.4,0.2],Polygon[{{+p,q},{+p,q+1},{+p+1,q+1},{+p+1,q}}]},{p,1,z},{q,0,Min[y,p]}]
},Background->Gray,ImageSize->{640,360},PlotRange->{{-7.5,8.5},{-0.5,8.5}}];
cutC[p_,x_,y_,z_]:=Block[{m1,m2,n,cx,cy,cz},
m1=p[[1]];m2=p[[2]];n=0.2;cx=0;cy=0;cz=0;
If[-x-1<m1<z+2&&-1<m2<y+2,
Do[If[-v-n<m1-0<-v+n,cx=x-v],{v,0,x}];
Do[If[+v-n<m1-1<+v+n,cz=z-v],{v,0,z}];
Do[If[+v-n<m2-1<+v+n,cy=y-v],{v,0,y}];
];
If[cx+cy+cz≠0&&(cx+cy==0||cy+cz==0||cz+cx==0),
tCount++;return={x-cx,y-cy,z-cz},
return={x,y,z}
];
return
];)

]

This research was published by Integers.

チョコレート問題において新しい問題をつくることはやさしい。形を変えるだけでいい。

It is easy to make a new chocolate games. You have only to change the shape.

Mathematica Code for chocolates

k=1;h=0;ss=40;al=Flatten[Table[{a,b,c},{a,0,2ss},{b,0,ss},{c,0,2ss}],2];
allcases=Select[al,(1/k)(#[[1]]+#[[3]]-h)≥#[[2]]&];

allcases//Length

move[z_]:=Block[{p},p=z;
Union[Table[{t1,Min[Floor[(1/k)(t1+p[[3]]+h)],p[[2]]],p[[3]]},{t1,0,p[[1]]-1}],
Table[{p[[1]],t2,p[[3]]},{t2,0,p[[2]]-1}],
Table[{p[[1]],Min[Floor[(1/k)(t3+p[[1]]+h)],p[[2]]],t3},{t3,0,p[[3]]-1}]
]
]

Mex[L_]:=Min[Complement[Range[0,Length[L]],L]];
Gr[pos_]:=Gr[pos]=Mex[Map[Gr,move[pos]]]

pposition=Select[allcases,Gr[#]==0&];

これらの方法で生徒の創造性を育てても、それは数学の創造性だけなのか?

When you cultivate creativity in students, is it only for mathematics?

答えは、「そうではない」

The answer is “No”.

新しいことを生み出すことは、数学以外のことでも、小さな変形によって可能です。これは、チェスとアメリカンフットボールを合わせたゲームで、アップル世界開発者会議でスカラーシップ(世界中で20才以下の140人の中に2人選ばれた）をとったもの。

With T.Cook

It is possible to create new things by changing some parts of the original. This is American Foot Ball Chess created by my student.

これは、普通紙の上で行う紙相撲をSmart Phoneに搭載したもの。これは、プレーする場所を変えた。

Paper Sumo Wrestling on Smartphone. This was created by changing the place where we play it.

私の次の目標は、Mathematicaを用いて高校生や大学学部生徒と共に新しい発見をする活動を広げて行くことです。
高校生に研究指導をするときに、Mathematicaを使わずに可能かということは、私にはわかりませんが、私の場合には、生徒との研究の99%以上はMathematicaなしには不可能だったものです。

My next aim is to promote math research for high school or undergraduate students using Mathematica in the world. I do not know what kind of math research is possible for these students if they do not use Mathematica. 99% of my result with the students were impossible without Mathematica.