x

y

A

B

O

阿良老師玩撞球，\;以球檯兩鄰邊為
\坐標軸且視球為點，\;如右圖所示。\;
\撞球檯的檯面是一個矩形，\;若將其
\放在坐標平面上，\;可設O為原點。
\若不考慮球的體積，\;將球視為一點，\;
\白球所在位置為A(2,4)，\;紅球所在
\位置為B(22,6)。\;已知白球從A(2,4)打
\出去，\;碰到檯邊的\;P\;點，\;再折向撞擊
\位於B(22,6)的紅球，\;試求\;P\;點坐標。

Forward

A(2,4)

B(22,6)

A'(2,-4)

P

O

∵∠BAO=30°
\⇒L的斜角為150°
\⇒m_L=tan150°=-\frac{1}{\sqrt{3}}

F = (14.86, -3.02)
s = 0 (slider from 0 to 26, step 1)
A = Corner(4)
B = (14.88, -7.38)
poly_0, b, a, c, d = Polygon(B, B + (18, 0), B + (18, 12), B + (0, 12))
C = PointIn(poly_0)
O = DynamicCoordinates(C, floor(x(C)), floor(y(C)))
u = Vector((x(B), y(O)), (x(B) + 18, y(O)))
v = Vector((x(O), y(B)), (x(O), y(B) + 12))
L_1: x - x(O) + (sqrt(3) * (y - y(O))) = 8
e_0 = OrthogonalLine(O, L_1)
H = Intersect(e_0, L_1)
P_A = Intersect(y = y(O), L_1)
P_B = Intersect(x = x(O), L_1)
e_1 = Segment(H, O)
α = Angle(O, H, P_A)
β = Angle(P_A, O, H)
e_2 = Segment(H, (x(H), y(O)))
t_H: "H" + (LaTeX[If[s ≥ 2, "(1,\sqrt{3})"]]) + ""
D: F + (18, 0)
O_1 = (16, -2.02)
E: F + (0, 9)
u_1 = Vector((x(E), y(O_1)), (x(D), y(O_1)))
v_1 = Vector((x(O_1), y(D)), (x(O_1), y(E)))
Q_A: O_1 + (1, 2)
Q_B: O_1 + (11, 3)
Q_C: O_1 + (1, (-2))
e = Segment(Q_C, Q_B)
Q_D: O_1 + (5, 0)
t = 2 (slider from 0 to 2, step 0.01)
R_A = Translate(Q_A, Vector[(t * Vector[Q_A, Q_D])])
R_B = Translate(Q_D, Vector[((t - 1) * Vector[Q_D, Q_B])])
t1 = 5 (slider from 0 to 5, step 0.01)
S_Q = Point(u_1)
f = Segment((x(S_Q), y(F)), (x(S_Q), y(E)))
G = Dilate(Mirror[Q_A, f], 2, S_Q)
g = Segment(Q_A, S_Q)
h = Segment(S_Q, G)
γ = Angle(Q_A, S_Q, S_Q + (0, 1))
δ = Angle(G, S_Q, S_Q + (0, 1))
i = Segment(Q_A, Q_C)
j = Segment(Q_A, Q_D)
I = (23.260000000000005, -3.02)
J = (17.580000000000005, -4.800000000000001)
K = (32.38000000000001, -4.800000000000001)
Question1: "坐標平面上，\;直線L分別交x軸與
\\y軸於A,B兩點，\;如圖所示。\;自原
\\點O作L的垂線\overline{OH}，\;其中H為垂
\\足；\;若\overline{OH}=2,∠AOH=60°，
\\求：\\" + (LaTeX[If[s  <  10, "(1)\;H的坐標與直線L的方程式；"]]) + "" + (LaTeX[If[s ≥ 10, "(2)\;線段\overline{AB}上，\;x坐標與y坐標 \\\;\;\;\;\;\;都是有理數的點共有幾個?"]]) + ""
text5: "L" + (LaTeX[If[s ≥ 9, ":x+\sqrt{3}y=4"]]) + ""
text19: "利用\fgcolor{yellow}{入射角等於反射角}\;" + (LaTeX[If[s ≥ 20, "\\考慮A點對x軸的對稱點A'(2,-4)"]]) + "" + (LaTeX[If[s ≥ 21, "\\則因為A',P,B三點共線，\; 所以\\P\;點在\overleftrightarrow{A'B}與x軸的交點上"]]) + "
\\" + (LaTeX[If[s ≥ 22, "\overleftrightarrow{A'B}:\frac{y+4}{x-2}=\frac{6+4}{22-2}\;"]]) + "" + (LaTeX[If[s ≥ 23, "\\⇒x-2y=10"]]) + "" + (LaTeX[If[s ≥ 24, "⇒將y=0代入，\;\\得x=10\;"]]) + "" + (LaTeX[If[s ≥ 25, "故得P(10,0)"]]) + ""
text11: "設(a,b)為\overline{AB}上的點，\;且a,b為
\\有理數
\\" + (LaTeX[If[s ≥ 12, "則a+\sqrt{3}b=4"]]) + "
\\" + (LaTeX[If[s ≥ 13, "若b≠0，\;"]]) + "" + (LaTeX[If[s ≥ 14, "則\sqrt{3}=\frac{4-a}{b}\fgcolor{yellow}{(不合)}"]]) + "
\\" + (LaTeX[If[s ≥ 15, "∴b=0"]]) + "" + (LaTeX[If[s ≥ 16, "⇒a=4"]]) + "
\\" + (LaTeX[If[s ≥ 17, "故\overline{AB}上x坐標與y坐標都是有理數\\的點只 有一個A(4,0)"]]) + ""
text1: "" + (LaTeX[If[(s ≥ 1) ∧ (s ≤ 7), "∵\overline{OH}=2,∠AOH=60°\;"]]) + "" + (LaTeX[If[(s ≥ 2) ∧ (s ≤ 7), "\\故H(1,\sqrt{3})"]]) + "
\\" + (LaTeX[If[(s ≥ 3) ∧ (s ≤ 7), "直線\overleftrightarrow{OH}的斜率\\m_{\overleftrightarrow{OH}}=tan60°"]]) + "" + (LaTeX[If[(s ≥ 4) ∧ (s ≤ 7), "=\sqrt{3}"]]) + "
\\" + (LaTeX[If[(s ≥ 5) ∧ (s ≤ 7), "又L⊥\overleftrightarrow{OH}"]]) + "" + (LaTeX[If[(s ≥ 6) ∧ (s ≤ 7), "⇒m_L×m_{\overleftrightarrow{OH}}=-1"]]) + "
\\" + (LaTeX[If[s ≥ 7, "⇒m_L=-\frac{1}{\sqrt{3}}"]]) + "
\\" + (LaTeX[If[s ≥ 8, "∴L:y-\sqrt{3}=-\frac{1}{\sqrt{3}}(x-1)"]]) + "
\\" + (LaTeX[If[s ≥ 9, "⇒L:x+\sqrt{3}y=4"]]) + ""