%
%	gbb60166@gmail.com	https://unilab.gbb60166.jp/prekou/prekou.htm
%
% aspectratio= は 1610, 169, 149, 54, 32 の中から選べる（省略時は 43）
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% navi. symbolsは目立たないが，dvipdfmxを使うと機能しないので非表示に
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\usepackage[deluxe]{otf}
%\usepackage{tikzsymbols}
%\usepackage[dvipdfmx]{graphicx}%
\def\pgfsysdriver{pgfsys-dvipdfmx.def}%（graphicxパッケージを使用しない場合はこの行を有効に）
%\def\pgfsysdriver{pgfsys-dvips.def}%デフォルト
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\nofiles

\hypersetup{% Jul 18 2023
	pdftitle={プレ高数学科（LaTeX Beamer で汎用PDFプレゼンを作る）},
	pdfauthor={gbb60166@gmail.com},
	pdfsubject={https://unilab.gbb60166.jp/prekou/prekou.htm}
	}

% フォントはお好みで
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% 参考にしたURL
% http://windom.phys.hirosaki-u.ac.jp/fswiki/wiki.cgi?page=LaTeX+Beamer%A4%C7%A5%D7%A5%EC%A5%BC%A5%F3%A5%C6%A1%BC%A5%B7%A5%E7%A5%F3

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\filldraw[color=CUDBlue, line width=1pt, rounded corners=2pt] (-0.1,0)--(2.1,0)--(2.1,1.1)--(-0.1,1.1)--cycle;
\draw(1,0.5) node[white]{#1};
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% ↑ 数式モードで式の前後に入るスペースを制御したかったが、
%    難しいのでやめた。段々難解なコードになっているのでやめた方がよい？
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%■等号付き不等号 by 美文書作成入門
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% カラーユニバーサルデザインを調べたつもりだがあまり自信がありません
% http://www.fukushihoken.metro.tokyo.jp/kiban/machizukuri/kanren/color.files/colorudguideline.pdf
% http://jfly.iam.u-tokyo.ac.jp/colorset/

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\definecolor{CUDYellow}{RGB}{255,241,0}
\definecolor{CUDBrown}{RGB}{128,64,0}

\definecolor{CUDOrange}{RGB}{246,170,0}

% ■ ベースカラー広い面積の塗り分けに用いる低・中彩度色
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\definecolor{CUDBrightGreen}{RGB}{119,217,168}
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\begin{document}

\title{プレ高数学科}\author{gbb60166}

%■■■■■■■■■■■■■　テスト領域　■■■■■■■■■■■■■■


%\end{document}
%■■■■■■■■■■■■■　完成品　■■■■■■■■■■■■■■

\begin{frame}[t]
\frametitle{例題}
折り紙だけを使って一番大きな正三角形をつくって下さい。

\bigskip
\small
\centerline{たけしのコマ大数学科 DVDBOX 第１期}

\centerline{１限目、問\ajMaru{4}、折り紙}

\bigskip
※\ 
\begin{minipage}[t]{0.9\textwidth}
正方形の折り紙の中に、最大の大きさの正三角形を折ってください
\end{minipage}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{この正三角形より大きなものが折れますよね…}
\vskip-3ex
\begin{tikzpicture}[scale=1,semithick, >=stealth,line join=round,
	gbb/.style={fill=white,inner sep=2pt},
	baseline={([yshift=0ex] current bounding box.north)}]
%\draw[help lines](-7.5,-3.5) grid (7.5,3.5);
\useasboundingbox(-7.5,-3.5) grid (7.5,3.5);
\scriptsize
\coordinate(A) at (-3,3);
\coordinate(B) at (-3,-3);
\coordinate(C) at (3,-3);
\coordinate(D) at (3,3);

\coordinate(AB) at ($ (A)!0.5!(B) $);
\coordinate(AD) at ($ (A)!0.5!(D) $);
\coordinate(BC) at ($ (B)!0.5!(C) $);
\coordinate(CD) at ($ (C)!0.5!(D) $);

\coordinate(P) at ($ ({3*sqrt(3)-3},0) $);
\coordinate(Q) at ($ (0,{3*sqrt(3)-3}) $);
%==================
% 点Bから線分PCに下ろした垂線の足
%\coordinate(BtoPC) at ($ (P)!(B)!(C) $);
%\path[name path=lineCD](C)--(D);
%\path[name path=lineX](B)--($ (B)!1.2!(BtoPC) $);
%% 点BからPCに下した垂線の足を延長して、CDと交わった点をRと定義する
%\path[name intersections={of=lineCD and lineX, by=R}];
%■ 最終的には∠RBC=15°になるので、こっちが楽だな
\coordinate(R) at ($ (3,{-3+6*tan(15)}) $);
%==================
\coordinate(S) at ($ ({-3+6*tan(15)},3) $);

\coordinate(DtoSR) at ($ (S)!(D)!(R) $);

\visible<1-2,4,6>{\filldraw[fill=CUDBrightCyan](B) rectangle (D);}
\visible<1>{\draw[CUDRed,very thick](B)--(C)--([turn] 120:6)--cycle;}
%%%+++++++++
\visible<3>{\filldraw[fill=CUDCream](AB) rectangle (D);
		\draw($ (A)!0.5!(CD) $) node{\LARGE 裏};
	}
%%%+++++++++
\visible<4>{\draw[dashed](AB)--(CD);}
%%%+++++++++
\visible<5>{\filldraw[fill=CUDCream](BC) rectangle (D);
		\draw($ (AD)!0.5!(C) $) node{\LARGE 裏};
		\draw[dashed]($ (AD)!0.5!(BC) $)--(CD);
	}
%%%+++++++++
\visible<6>{\draw[dashed](AB)--(CD);
	\draw[dashed](AD)--(BC);
	}
%%%+++++++++
\visible<7>{\filldraw[fill=CUDBrightCyan](A)--(B)--(R)--(D)--cycle;
	\draw[dashed](AD)--(0,-2);
	\filldraw[fill=CUDCream](B)--(R)--(P)--cycle;
	\draw[dashed](AB)--(CD);
	\draw[CUDRed,very thick](AB)-- node[gbb]{\(1\)}(B)
			-- node[gbb]{\(2\)}(P)
			-- node[gbb]{\(\sqrt{3}\)} cycle;
	\draw[dashed](B)--(C)--(R);
	\pic["\tiny\( 60^\circ \)",draw,angle eccentricity=1.5,
		angle radius=4ex]{angle=P--B--AB};
	\pic["\tiny\( 15^\circ \)",draw,angle eccentricity=1.5,
		angle radius=6ex]{angle=R--B--P};
	\pic["\tiny\( 15^\circ \)",draw,angle eccentricity=1.5,
		angle radius=8ex]{angle=C--B--R};
	}
%%+++++++++
\visible<8>{\filldraw[fill=CUDBrightCyan](D)--(S)--(B)--(R)--(D)--cycle;
	\draw[dashed](AD)--(0,-2)
			(CD)--(-2,0);
	\filldraw[fill=CUDCream](B)--(P)--(R)--cycle
				(B)--(Q)--(S)--cycle;
	\draw[dashed](B) rectangle (D);
	\pic["\tiny\( 60^\circ \)",draw,angle eccentricity=1.5,
		angle radius=6ex]{angle=R--B--S};
	\pic["\tiny\( 15^\circ \)",draw,angle eccentricity=1.5,
		angle radius=8ex]{angle=S--B--A};
	\pic["\tiny\( 15^\circ \)",draw,angle eccentricity=1.5,
		angle radius=8ex]{angle=C--B--R};
	}
%%%+++++++++
\visible<9>{\filldraw[fill=CUDBrightCyan](S)--(B)--(R)--cycle;
	\filldraw[fill=CUDCream](B)--($ (B)!0.6!(P) $)--(R)--cycle
				(B)--($ (B)!0.6!(Q) $)--(S)--cycle;
	\filldraw[fill=CUDCream]($ (D)!2!(DtoSR) $)--(S)--(R)--cycle;
	\draw[CUDRed,very thick](B)--(R)--(S)--cycle;

	\draw[CUDRed] let \p1=(B), \p2=(R), \p3=(S) in
		($ (B)!0.5!(R) $)
			node[rotate=atan((\x2-\x1)/(\y1-\y2))]{\large\(=\)}
		($ (B)!0.5!(S) $)
			node[rotate=atan((\x3-\x1)/(\y1-\y3))]{\large\(=\)};

	\draw[dashed](B) rectangle (D);
	\pic["\tiny\( 60^\circ \)",draw,angle eccentricity=1.5,
		angle radius=6ex]{angle=R--B--S};
	\pic["\tiny\( 15^\circ \)",draw,angle eccentricity=1.5,
		angle radius=8ex]{angle=S--B--A};
	\pic["\tiny\( 15^\circ \)",draw,angle eccentricity=1.5,
		angle radius=8ex]{angle=C--B--R};

	\pic["\tiny\( 60^\circ \)",draw,angle eccentricity=1.5,
		angle radius=6ex]{angle=S--R--B};
	\pic["\tiny\( 60^\circ \)",draw,angle eccentricity=1.5,
		angle radius=6ex]{angle=B--S--R};
	}
%■デバッグ用
%\foreach \z in {A,B,C,D,AB,AD,BC,CD,P,Q,R,S,DtoSR}
%	\draw(\z) node{\fontsize{6}{6}\selectfont\textrm{\z}};
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{では本題です}
折り紙で折れる最大の正六角形を作りなさい
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{さっきのやり方で正三角形を作って…}
\vskip-3ex
\begin{tikzpicture}[scale=1,semithick, >=stealth,line join=round,
	gbb/.style={fill=white,inner sep=2pt},
	baseline={([yshift=0ex] current bounding box.north)}]
%\draw[help lines](-7.5,-3.5) grid (7.5,3.5);
\useasboundingbox(-7.5,-3.5) grid (7.5,3.5);

\coordinate(O) at (0,0);
\coordinate(A) at (-3,3);
\coordinate(B) at (-3,-3);
\coordinate(C) at (3,-3);
\coordinate(D) at (3,3);

\coordinate(AB) at ($ (A)!0.5!(B) $);
\coordinate(AD) at ($ (A)!0.5!(D) $);
\coordinate(BC) at ($ (B)!0.5!(C) $);
\coordinate(CD) at ($ (C)!0.5!(D) $);

\coordinate(R) at ($ (3,{3*tan(15)}) $);
\coordinate(S) at ($ (O)!1!60:(R) $);%点Rを、点Oを中心として60°回転
%正確に言うなら「線分ORの1:0の内分点を、点Oを中心として60°回転」
% pgfmanual.pdf の150ページにこの書き方がある
\coordinate(T) at ($ (O)!1!120:(R) $);
\coordinate(U) at ($ (O)!1!180:(R) $);
\coordinate(V) at ($ (O)!1!240:(R) $);
\coordinate(W) at ($ (O)!1!300:(R) $);

% 点Cから線分SVに下ろした垂線の足
\coordinate(L) at ($ (S)!(C)!(V) $);
% 点Aから線分SVに下ろした垂線の足
\coordinate(M) at ($ (S)!(A)!(V) $);
% 点Cから線分VWに下ろした垂線の足
\coordinate(N) at ($ (V)!(C)!(W) $);
\coordinate(P) at ($ (3,{-3+(3*tan(15)+3)*tan(15)}) $);
% 点Pから線分RWに下ろした垂線の足
\coordinate(Q) at ($ (R)!(P)!(W) $);
% 点Aから線分STに下ろした垂線の足
\coordinate(E) at ($ (S)!(A)!(T) $);
\coordinate(F) at ($ (-3,{3-(3*tan(15)+3)*tan(15)}) $);
% 点Fから線分TUに下ろした垂線の足
\coordinate(G) at ($ (T)!(F)!(U) $);
% 点Aから線分TUに下ろした垂線の足
\coordinate(X) at ($ (T)!(A)!(U) $);
\coordinate(X') at ($ ({-3+(3*tan(15)+3)*tan(15)},3) $);
% 点Cから線分RWに下ろした垂線の足
\coordinate(Y) at ($ (R)!(C)!(W) $);
\coordinate(Y') at ($ ({3-(3*tan(15)+3)*tan(15)},-3) $);
%==================

\visible<1>{\filldraw[fill=CUDBrightCyan](A)--(B)--(C)--(D)--cycle;
	\draw[dashed](AB)--(CD)
		(BC)--(AD);
	}
%%%+++++++++
\visible<2-3>{\filldraw[fill=CUDBrightCyan]
			(A)--(B)--(C)--(R)--(S)--cycle;
	\filldraw[fill=CUDCream](S)|-(R)--cycle;
	\draw[dashed](AB)--(CD)
		(BC)--(AD)
		(S)--(O)--(R);
	}
%%%+++++++++
\visible<3>{\draw[dashed](U)--(O)--(V);}
%%%%+++++++++
\visible<4-5>{\filldraw[fill=CUDBrightCyan]
		(A)--(U)--(V)--(C)--(R)--(S)--cycle;
	\filldraw[fill=CUDCream](S)|-(R)--cycle
				(U)-|(V)--cycle;
	\draw[dashed](AB)--(CD)
		(BC)--(AD)
		(S)--(O)--(R)
		(U)--(O)--(V);
	}
%%%%+++++++++
\visible<5>{\draw[dashed](A)--(C);}
%%%%+++++++++
\visible<6>{\filldraw[fill=CUDBrightCyan](A)--(AB)--(S)--cycle;
	\draw[dashed](A)--(O);
	\filldraw[fill=CUDCream]($ (C)!2!(L) $)--(V)--(S)--(T)--cycle;
	}
%%%%+++++++++
\visible<7>{\filldraw[fill=CUDBrightCyan](S)--(F)--(U)--cycle;
	%\draw[dashed](A)--(O);
	\filldraw[fill=CUDCream]($ (C)!2!(L) $)--(V)--(S)--(T)--cycle;
	\filldraw[fill=CUDCream]($ (A)!2!(E) $)--(F)--(S)--cycle;
	}
%%%%+++++++++
\visible<8>{\filldraw[fill=CUDBrightCyan]
			(A)--(U)--(V)--(C)--(R)--(S)--cycle;
	\filldraw[fill=CUDCream](S)|-(R)--cycle
				(U)-|(V)--cycle;
	\draw[dashed](AB)--(CD)
		(BC)--(AD)
		(S)--(F)
		(S)--(O)--(R)
		(U)--(O)--(V)
		(A)--(O)
		(C)--(O);
	\fill[CUDRed](T) circle(1.6pt);
	}
%%%+++++++++
\visible<9>{\filldraw[fill=CUDBrightCyan](O)--(V)--(C)--(CD)--cycle;
	\draw[dashed](C)--(O);
	\filldraw[fill=CUDCream]($ (A)!2!(M) $)--(W)--(V)--(S)--cycle;
	}
%%%%+++++++++
\visible<10>{\filldraw[fill=CUDBrightCyan](O)--(V)--(P)--(CD)--cycle;
	%\draw[dashed](C)--(O);
	\filldraw[fill=CUDCream]($ (A)!2!(M) $)--(W)--(V)--(S)--cycle;
	\filldraw[fill=CUDCream]($ (C)!2!(N) $)--(P)--(V)--cycle;
	}
%%%%+++++++++
\visible<11>{\filldraw[fill=CUDBrightCyan]
			(A)--(U)--(V)--(C)--(R)--(S)--cycle;
	\filldraw[fill=CUDCream](S)|-(R)--cycle
				(U)-|(V)--cycle;
	\draw[dashed]
		(O)--(AB)
		(O)--(CD)
		(O)--(BC)
		(O)--(AD)
		(S)--(F)
		(P)--(V)
		(S)--(O)--(R)
		(U)--(O)--(V)
		(O)--(A)
		(O)--(C);
	\fill[CUDRed](T) circle(1.6pt)
			(W) circle(1.6pt);
	}
%%%%+++++++++
\visible<12>{\filldraw[fill=CUDBrightCyan]
			(X')--(U)--(V)--(Y')--(R)--(S)--cycle;
	\filldraw[fill=CUDCream](S)|-(R)--cycle
				(U)-|(V)--cycle;
	\draw[dashed]
		(O)--($ (O)!0.92!(AB) $)
		(O)--($ (O)!0.92!(CD) $)
		(O)--(BC)
		(O)--(AD)
		(S)--(T)
		(W)--(V)
		(O)--($ (A)!0.2!(C) $)
		(O)--($ (A)!0.8!(C) $)
		(S)--(O)--(R)
		(U)--(O)--(V);
	\filldraw[fill=CUDCream]($ (A)!2!(X) $)--(X')--(U)--cycle;
	\filldraw[fill=CUDCream]($ (C)!2!(Y) $)--(Y')--(R)--cycle;
	\fill[CUDRed](T) circle(1.6pt)
			(W) circle(1.6pt);
	}
%%%%+++++++++
\visible<13>{\filldraw[fill=CUDBrightCyan]
			(T)--(U)--(V)--(W)--(R)--(S)--cycle;
	\filldraw[fill=CUDCream](S)|-(R)--cycle
				(U)-|(V)--cycle;
	\draw[dashed]
		(O)--($ (O)!0.92!(AB) $)
		(O)--($ (O)!0.92!(CD) $)
		(O)--($ (O)!0.92!(BC) $)
		(O)--($ (O)!0.92!(AD) $)
		(S)--(O)--(R)
		(U)--(O)--(V)
		(S)--(T)--(U)
		(O)--($ (A)!0.2!(C) $)
		(O)--($ (A)!0.8!(C) $);
	\filldraw[fill=CUDCream]($ (C)!2!(N) $)--(W)--(V)--cycle;
	\filldraw[fill=CUDCream]($ (P)!2!(Q) $)--(W)--(R)--cycle;
	\filldraw[fill=CUDCream]($ (A)!2!(E) $)--(T)--(S)--cycle;
	\filldraw[fill=CUDCream]($ (F)!2!(G) $)--(T)--(U)--cycle;
	}
%%%%+++++++++
%■デバッグ用
%\foreach \z in {O,A,B,C,D,AB,AD,BC,CD,R,S,T,U,V,W,X,X',Y,Y',L,M,N,P,Q,E,F,G}
%	\draw(\z) node{\fontsize{6}{6}\selectfont\textrm{\z}};
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{美しき数学の時間}
番組の中では正七角形や正十三角形の折り紙も紹介されていました。

\bigskip
定規とコンパスだけでは三次方程式を解けませんが、
折り紙では\textcolor{CUDRed}{三次元折り}と呼ばれる折り方をすることに
よって、定規とコンパスだけでは作図できない点も折ることができるそうです。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{折り紙の三次元折}
\vskip-3ex
\begin{tikzpicture}[scale=1,thick, >=stealth,line join=round,
	gbb/.style={fill=white,inner sep=2pt},
	baseline={([yshift=0ex] current bounding box.north)}]
%\draw[help lines](0,0) grid (15,7);
\useasboundingbox(0,0) grid (15,7);
\scriptsize
\coordinate[label=above left:{\textrm{A}}](A) at (4,5);
\coordinate[label=below right:{\textrm{C}}](C) at (11,1);
\coordinate[label=above:{\textrm{E}}](E) at (6,5);

\path let \p1=(A), \p2=(C) in
	coordinate[label=below left:{\textrm{B}}](B) at (\x1,\y2)
	coordinate[label=above right:{\textrm{D}}](D) at (\x2,\y1);

\coordinate[label=left:{\textrm{F}}](F) at ($ (A)!1/4!(B) $);
\coordinate[label=right:{\textrm{G}}](G) at ($ (D)!1/4!(C) $);
\coordinate[label=left:{\textrm{H}}](H) at ($ (A)!3/4!(B) $);
\coordinate[label=right:{\textrm{I}}](I) at ($ (D)!3/4!(C) $);

\draw(7.5,6.5) node{\small 【\textrm{B}が\textrm{HI}の上に】
	《\textrm{F}が\textrm{BE}の上に》同時にのるように折る};

\visible<1>{\filldraw[fill=CUDBrightCyan](A) rectangle (C);
	\draw[dashed,name path=lineFG](F)--(G);
	\draw[dashed,name path=lineHI](H)--(I);
	\draw[dashed,name path=lineBE](B)--(E);
	}

\path[name path=lineAD](A)--(D);
\path[name path=lineBC](B)--(C);
\path[name path=lineX](5,6)--++(3,-6);

\path[name intersections={of=lineBE and lineX, by=P}];
\path[name intersections={of=lineHI and lineX, by=Q}];

\coordinate(P') at ($ (F)!0.5!(P) $);
\coordinate(Q') at ($ (B)!0.5!(Q) $);

\path[name path=lineP'Q']($ (Q')!1.5!(P') $)--($ (P')!1.5!(Q') $);

\path[name intersections={of=lineP'Q' and lineAD, by=S}];
\path[name intersections={of=lineP'Q' and lineBC, by=T}];

\coordinate(U) at ($ (S)!(A)!(T) $);

%\draw(T)--(Q)--($ (A)!2!(U) $)--(S)--cycle;
\visible<2>{\filldraw[fill=CUDBrightCyan](S)--(T)--(C)--(D)--cycle;
	\draw[dashed](F)--(G)
		(H)--(I)
		(B)--(E)
		(S)--(A)--(B)--(T);
	\filldraw[fill=CUDCream](S)--(T)--(Q)--($ (A)!2!(U) $)--cycle;
	\fill[CUDRed](P) circle(2.4pt);
	\filldraw[ultra thick,fill=white,draw=CUDBlue](Q) circle(2.4pt);
	}

\filldraw[ultra thick,fill=white,draw=CUDBlue](B) circle(2.4pt);
\fill[CUDRed](F) circle(2.4pt);
%■デバッグ用
%\foreach \z in {P,Q,S,T,P',Q'}
%	\draw(\z) node{\fontsize{6}{6}\selectfont\textrm{\z}};
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{美しき数学の時間}
三次元折りは、$2$つの放物線の共通の接線を求めることになるそうです。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\end{document}

