STAINED GLASS ART FUNCTIONS
&
“Jean Monnet” High School - Bucharest, Romania Gaudeamus High School - Chisinău, Republic of Moldova Primary School Stevan Dukic - Belgrade, Serbia
Mihaela Git, Ludmila Cojocari, Katarina Ivanovic, Finica Rotaru 2019-2020
Aceste probleme au fost propuse pe baza desenelor inspirate din vitraliile văzute de elevii in viața reală. Rezolvările au fost lucrate de elevi, fie individual, fie in grupe naționale sau internaționale. Au participat: clasele a VII-a, a VIII-a, a IV-a din Liceul Jean Monnet, Liceul Gaudeamus și Primary School Stevan Dukic.
These problems were proposed based on the drawings inspired by the stained glass windows seen by the students in real life. The solutions were worked by students, either individually, or in national or international groups. th 7th, 8 , 4th grades from Jean Monnet High School, Gaudeamus High School and Primary School Stevan Dukic participated.
Enunțuri/Enunciations 1. In this figure ABCD is a square and A(–2; –1), G(–2; 0), I(0; 1). a) Find the coordinate for H b) Find the functions determined by the lines AK, BH, GI. c) * Find the functions determined by the lines KJ, EH.
2. The adjoining figure is a stained glass window. If the points have the coordinates: A(0; –4), B(5; –4), C(5; 4), F(2; 4), I(4.5; 2.2), M(0; 2), N(5; –2), find the functions determined by the lines: AB, AN, CM, ML, IF.
3. Find the functions determinated by the points: 1) A and C 2) B and D 3) E and G 4) E and F 5) E and H 6) E and C 7) B and G 8) C and D 9) B and H. The coordinates of the points are in the attached figure.
4. Find the functions determinated by the lines: 1) DE, with D(0;1.6), E(3; 2) 2) FG, with F(0; 8), G(3; 0) 3) HI, H(0; –1,6), I(1,2; –3) 4) JK, J(0; 0,8), K(3; 2) 5) LM, L(0; 0.5), M(3; 0)
5. Find the functions determinated by the lines: 1) AB, with A(0; 1), B(1; 2) 2) CD, with C(0; –2), D(2; –1) 3) EF, E(0; 1,4), F(2; 0) 4) GH, G(0; –1), H(–2;1)
6. Find the function determinated by the line AB, if: 1) A(0; 0), B(1.2; 2) 2) C(-4; -3), D(0; 1,4) 3) G(0; –1,6), H(–3; 1.4) 4) I(–5; 0.4), J(–3.2; –2) 5) K(–5; –1.5), L(3; 0.3)
7. Find the function determinated by the line AD, BC, KL, RS, if: A(0; –4), B(–3; 2), C(0; -3), D(-2; 2) E(–1; 2), F(0; –2), G(0; 2), H(1; –2)
8. Find the function determinated by the line: GH, LN, LK, JI, MN, if: G(–3; 1), H(0; –1), I(1; 1), J(0; –2), K(–1; 1), L(2; –1), M(2; 0),N(–2; –1)
9. Find the functions determinated by the line IO, KJ, ML, GH, if: G(–1.5; 5.5) , I(–0.76; 7.6), J(4;0); H(6.3, 5.5) , K(4.4; 7.6), L(–1.4; 2) , M(6.3; 2), O(0; 0).
10. In this stained glass the coordinates of the points are given, thus: A(–6; –6), B(0;–6), C(0;0), E(– 2; –2), I(–10; 2), L(–5; 3). Determine the functions described by the graphs of the rights AC, AE, CI, CL
11. Find the function determinated by the lines:AB, AC, AE,CD, if: A(0; 3), B(2.86; 0.34) C(2.38; 5), D(1; 6.58) E(1.3; 2.9)
SOLUTIONS 1.
a) H(3;0). b) 𝑓(𝑥) = a𝑥 + b, a, b ∈ R • K(−0.2; 0) ∈ 𝐺𝑓 ⇒ 𝑓(−0.2) = 0 ⇒ −0.2𝑎 + 𝑏 = 0 (1) 𝐴(−2; −1) ∈ 𝐺𝑓 ⇒ 𝑓(−2) = −1 ⇒ −2𝑎 + 𝑏 = −1 (2) (2)– (1) ⇒ −1.8𝑎 = −1 ⇒ −1 10 5 𝑎= = = −1.8 18 9 2 5 1 𝑏 = 0.2𝑎 = ∙ = 10 9 9 5 𝟏 𝒇: 𝑹 ⟶ 𝑹, 𝒇(𝒙) = 𝒂 + , • B(2; – 1) ∈ 𝐺𝑓 ⇒ 𝑓(2) = −1 ⇒ 2𝑎 + 9 𝟗 𝑏 = −1 (1) • G(−2; 0) ∈ 𝐺𝑓 ⇒ 𝑓(−2) = 0 ⇒ 𝐻(1; 3) ∈ 𝐺𝑓 ⇒ 𝑓(1) = 3 ⇒ 𝑎 + 𝑏 = 3 (2) −2𝑎 + 𝑏 = 0 (1) 𝐼(0; 1) ∈ 𝐺𝑓 ⇒ 𝑓(0) = 1 ⇒ 𝑏 = 1(2) (1)– (2) ⇒ 𝑎 = −4 ⇒ 𝑏 = 7 1 𝒇: 𝑹 ⟶ 𝑹, 𝒇(𝒙) = −𝟒𝒙 + 𝟕 (1)& (2) ⇒ 𝑎 = 2 𝒙 𝒇: 𝑹 ⟶ 𝑹, 𝒇(𝒙) = + 𝟏 𝟐 2.
𝑓: 𝑅 → 𝑅, 𝑓(𝑥) = 𝑎𝑥 + 𝑏, 𝑎, 𝑏 ∈ 𝑅 AB: AB||Ox, y= –4 AN: 𝐴, 𝑁 ∈ 𝐺𝑓 ⇒ 𝑓(0) = −4, 𝑓(5) = −2 ⇒ 𝑏 = −4, 5𝑎 + 𝑏 = −2 ⇒ 2 5𝑎 − 4 = −2 ⇒ 𝑎 = ⇒ 5 𝟐 𝒇(𝒙) = 𝒙 − 𝟒 𝟓 CM: 𝑀, 𝐶 ∈ 𝐺𝑓 ⇒ 𝑓(0) = 2, 𝑓(5) = 4⇒ 𝑏 = 2, 5𝑎 + 𝑏 = −2 ⇒ 5𝑎 − 4 = 4 8 ⇒𝑎= ⇒ 5 𝟖 𝒇(𝒙) = 𝒙 + 𝟐 𝟓
3.
𝑓(𝑥) = a𝑥 + b, a, b ∈ R A(– 2; – 3) ∈ 𝐺𝑓 ⇒ 𝑓(−2) = −3 ⇒ −2𝑎 + 𝑏 = −3 (1) 𝐶(2; 4) ∈ 𝐺𝑓 ⇒ 𝑓(2) = 4 ⇒ 2𝑎 + 𝑏 = 4 (2) (1)+ (2) ⇒ 2𝑏 = 1 ⇒ 𝑏 = 0.5 (2) ⇒ 2𝑎 + 0.5 = 4 ⇒ 2𝑎 = 3.5 ⇒ 𝑎 = 1.75 𝒇(𝒙) = 𝟏. 𝟕𝟓𝒂 + 𝟎. 𝟓, 𝒇: 𝑹 ⟶ 𝑹 2. D, C symmetrical points ⇒ 𝐷(−2; 4) 𝑓(𝑥) = a𝑥 + b, a, b ∈ R D(– 2; 4) ∈ 𝐺𝑓 ⇒ 𝑓(−2) = 4 ⇒ −2𝑎 + 𝑏 = 4 (1) 𝐵(2; −3) ∈ 𝐺𝑓 ⇒ 𝑓(2) = −3 ⇒ 2𝑎 + 𝑏 = −3 (2) (1)+ (2) ⇒ 2𝑏 = 1 ⇒ 𝑏 = 0.5 (2) ⇒ 2𝑎 + 0.5 = −3 ⇒ 2𝑎 = −3.5 ⇒ 𝑎 = −1.75 𝒇(𝒙) = 𝟏. 𝟕𝟓𝒂 + 𝟎. 𝟓, 𝒇: 𝑹 ⟶ 𝑹 9. 𝑓(𝑥) = a𝑥 + b, a, b ∈ R H(1; 2.25) ∈ 𝐺𝑓 ⇒ 𝑓(1) = 2.25 ⇒ 𝑎 + 𝑏 = 2.25 (1) 𝐵(2; −3) ∈ 𝐺𝑓 ⇒ 𝑓(2) = −3 ⇒ 2𝑎 + 𝑏 = −3 (2) (1)– (2) ⇒ 2𝑏 = 1 ⇒ −𝑎 = 5.25 ⇒ 𝑎 = −5.25 (1) ⇒ 𝑏 = 2.25 − (−5.25) = 7.5 𝒇(𝒙) = −𝟓. 𝟐𝟓𝒂 + 𝟕. 𝟓, 𝒇: 𝑹 ⟶ 𝑹
4.
5.
6.
𝑓: 𝑹 ⟶ 𝑹, 𝑓(𝑥) = a𝑥 + b, a, b ∈ R 1) A(0; 0), B(1,2; 2) 𝐴(0; 0) ∈ 𝐺𝑓 ⇒ 𝑓(0) = 0 ⇒ 𝑏 = 0 B(1.2; 2) ∈ 𝐺𝑓 ⇒ 𝑓(1.2) = 2 ⇒ 1.2𝑎 = 2 ⇒ 𝑎 = 0. (6) 𝒇: 𝑹 ⟶ 𝑹, 𝒇(𝒙) = 𝟎. (𝟔)𝒙 2) C(–4; –3), D(0; 1,4) 𝐷(0; 1.4) ∈ 𝐺𝑓 ⇒ 𝑓(0) = 1.4 ⇒ 𝑏 = 1.4 C(–4; –3) ∈ 𝐺𝑓 ⇒ 𝑓(−4) = −3 ⇒ −4𝑎 + 1.4 = −3 ⇒ −4𝑎 = −4.4 ⇒ 𝑎 = 1.1 𝒇: 𝑹 ⟶ 𝑹, 𝒇(𝒙) = 𝟏. 𝟏𝒙 + 𝟏. 𝟒 3) G(0; –1.6), H(–3; 1.4) 𝐺(0; −1.6) ∈ 𝐺𝑓 ⇒ 𝑓(0) = −1.6 ⇒ 𝑏 = −1.6 G(–3; –1.4) ∈ 𝐺𝑓 ⇒ 𝑓(−3) = 1.4 ⇒ −3𝑎 − 1.6 = 1.4 ⇒ −3𝑎 = 3 ⇒ 𝑎 = −1 𝒇: 𝑹 ⟶ 𝑹, 𝒇(𝒙) = −𝒙 − 𝟏. 𝟔 4) I(–5; 0,4), J(–3.2; –2) −5𝑎 + 𝑏 = 0.4; −3.2𝑎 + 𝑏 = −2 1.8𝑎 = −2.4 ⇒ 𝑎 = −1. (3) ⇒ 𝑏 = 6.27 5) K(–3; –1), L(3; 0,2) −5𝑎 + 𝑏 = −1.5, 3𝑎 + 𝑏 = 0.3 ⇒ 8𝑎 = 1.8 ⇒ 𝑎 = 0.225 ⇒ 𝑏 = 0.3 − 0.675 = −0.375 𝒇: 𝑹 ⟶ 𝑹, 𝒇(𝒙) = −𝟎. 𝟐𝟐𝟓𝒙 − 𝟎. 𝟑𝟕𝟓
7.
8.
9. 𝑓: 𝑹 ⟶ 𝑹, 𝑓(𝑥) = a𝑥 + b, a, b ∈ R • 𝑂(0; 0) ∈ 𝐺𝑓 ⇒ 𝑓(0) = 0 ⇒ 𝑏 = 0 (1) I(–0.76; 7.6) ∈ 𝐺𝑓 ⇒ 𝑓(– 0.76) = 7.6 ⇒ – 0.76𝑎 + 𝑏 = 7.6 (2) (1), (2) ⇒ 𝑎 = −10 𝒇(𝒙) = −𝟏𝟎𝒙, 𝒇: 𝑹 ⟶ 𝑹 • K(4.4; 7.62) ∈ 𝐺𝑓 ⇒ 𝑓(4.4) = 7.62 ⇒ 4.4𝑎 + 𝑏 = 7.62 (1) J(4;0) ∈ 𝐺𝑓 ⇒ 𝑓(4) = 0 ⇒ 4𝑎 + 𝑏 = 0 The equations are subtracted, member by member ⇒ 0.4𝑎 = 7.62 ⇒ 𝑎 = 19.05 ⇒ 𝑏 = −4𝑎 ⇒ 𝑏 = −76.1 𝑓: 𝑹 ⟶ 𝑹, 𝑓(𝑥) = 19.05𝑥 − 76.1 • G(–1.5; 5.5) , H(6.3, 5.5) y=constant ⇒ 𝑦 = 5.5 • L(–1.4; 2) , M(6.3; 2) y=constant ⇒ 𝑦 = 2
10. 𝑓: 𝑹 ⟶ 𝑹, 𝑓(𝑥) = a𝑥 + b, a, b ∈ R GeoGebra solution: AE: y=x ⇒ f(x)=x CE: y=2.13x ⇒ f(x)=2.13x CL: y= –0.6x ⇒ f(x)= – 0.6x −𝑥 CI: x+5y=0 ⇒ f(x)= 5 AB: y= –6 ⇒ f(x)= – 6x
11.
GeoGebra solution: đ?‘“: đ?‘š â&#x;ś đ?‘š, đ?‘“(đ?‘Ľ) = ađ?‘Ľ + b, a, b ∈ R
AB: f(x)=-0.93x+3 AC: f(x)=-0.84x+3 CD: f(x)=--1.14x+7.72 AE: f(x)=-0.08x+3