Olympiad Maths Prep

Track / Stage 8 / 87 of 180 #1787 of 2000

Problem 1787

IMO Shortlist mid-range; USAMO P2/P5
Geometry Difficulty 8.3 Prove it

Consider four points P1, P2, P3P_1,\ P_2,\ P_3 and P4P_4 on the coordinate plane with origin OO which satisfy the condition OP→n−1+OP→n+1=32OP→n, (n=2,3). \overrightarrow {{OP}}_{n-1}+\overrightarrow {{OP}}_{n+1}=\frac{3}{2} \overrightarrow {{OP}}_{n},\ (n=2,3).

(1) If P1, P2P_1,\ P_2 lie on the curve xy=1,xy=1, then prove that P3P_3 doesn't lie on this curve.

(2) If P1, P2P_1,\ P_2 and P3P_3 lie on the circle x2+y2=1,x^2+y^2=1, then prove that P4P_4 lie on this circle as well.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

### Part (1)
1. Given the points P1 P_1 and P2 P_2 lie on the curve xy=1 xy = 1 , we can represent them as:
OP1→=⟨a,1a⟩andOP2→=⟨b,1b⟩ \overrightarrow{OP_1} = \langle a, \frac{1}{a} \rangle \quad \text{and} \quad \overrightarrow{OP_2} = \langle b, \frac{1}{b} \rangle
where a≠0 a \neq 0 and b≠0 b \neq 0 .

2. Using the given condition OP1→+OP3→=32OP2→ \overrightarrow{OP_{1}} + \overrightarrow{OP_{3}} = \frac{3}{2} \overrightarrow{OP_{2}} , we can solve for OP3→ \overrightarrow{OP_3} :
OP3→=32OP2→−OP1→=⟨32b−a,32b−1a⟩ \overrightarrow{OP_3} = \frac{3}{2} \overrightarrow{OP_2} - \overrightarrow{OP_1} = \left\langle \frac{3}{2}b - a, \frac{3}{2b} - \frac{1}{a} \right\rangle

3. For P3 P_3 to lie on the curve xy=1 xy = 1 , the coordinates of OP3→ \overrightarrow{OP_3} must satisfy:
(32b−a)(32b−1a)=1 \left( \frac{3}{2}b - a \right) \left( \frac{3}{2b} - \frac{1}{a} \right) = 1

4. Simplifying the left-hand side:
(32b−a)(32b−1a)=32⋅32b−32⋅1a−a⋅32b+a⋅1a \left( \frac{3}{2}b - a \right) \left( \frac{3}{2b} - \frac{1}{a} \right) = \frac{3}{2} \cdot \frac{3}{2b} - \frac{3}{2} \cdot \frac{1}{a} - a \cdot \frac{3}{2b} + a \cdot \frac{1}{a}
=94b−32a−3a2b+1 = \frac{9}{4b} - \frac{3}{2a} - \frac{3a}{2b} + 1

5. Setting the equation equal to 1:
94b−32a−3a2b+1=1 \frac{9}{4b} - \frac{3}{2a} - \frac{3a}{2b} + 1 = 1
94b−32a−3a2b=0 \frac{9}{4b} - \frac{3}{2a} - \frac{3a}{2b} = 0
94b−3a2b=32a \frac{9}{4b} - \frac{3a}{2b} = \frac{3}{2a}
9−6a4b=32a \frac{9 - 6a}{4b} = \frac{3}{2a}
2a(9−6a)=12b 2a(9 - 6a) = 12b
18a−12a2=12b 18a - 12a^2 = 12b
3a−2a2=2b 3a - 2a^2 = 2b

6. This is a quadratic equation in terms of a a and b b . The discriminant of this quadratic equation is:
Δ=(−3)2−4⋅2⋅2=9−16=−7 \Delta = (-3)^2 - 4 \cdot 2 \cdot 2 = 9 - 16 = -7
Since the discriminant is negative, there are no real solutions for a a and b b . Therefore, P3 P_3 cannot lie on the curve xy=1 xy = 1 .

### Part (2)
1. Given P1,P2, P_1, P_2, and P3 P_3 lie on the circle x2+y2=1 x^2 + y^2 = 1 , we can represent them as:
OP1→=⟨cos⁡α,sin⁡α⟩,OP2→=⟨cos⁡β,sin⁡β⟩ \overrightarrow{OP_1} = \langle \cos \alpha, \sin \alpha \rangle, \quad \overrightarrow{OP_2} = \langle \cos \beta, \sin \beta \rangle

2. Using the given condition OP1→+OP3→=32OP2→ \overrightarrow{OP_{1}} + \overrightarrow{OP_{3}} = \frac{3}{2} \overrightarrow{OP_{2}} , we can solve for OP3→ \overrightarrow{OP_3} :
OP3→=32OP2→−OP1→=⟨32cos⁡β−cos⁡α,32sin⁡β−sin⁡α⟩ \overrightarrow{OP_3} = \frac{3}{2} \overrightarrow{OP_2} - \overrightarrow{OP_1} = \left\langle \frac{3}{2} \cos \beta - \cos \alpha, \frac{3}{2} \sin \beta - \sin \alpha \right\rangle

3. Since P3 P_3 lies on the circle x2+y2=1 x^2 + y^2 = 1 , we have:
(32cos⁡β−cos⁡α)2+(32sin⁡β−sin⁡α)2=1 \left( \frac{3}{2} \cos \beta - \cos \alpha \right)^2 + \left( \frac{3}{2} \sin \beta - \sin \alpha \right)^2 = 1

4. Expanding and simplifying:
(32cos⁡β−cos⁡α)2+(32sin⁡β−sin⁡α)2=(94cos⁡2β−3cos⁡αcos⁡β+cos⁡2α)+(94sin⁡2β−3sin⁡αsin⁡β+sin⁡2α) \left( \frac{3}{2} \cos \beta - \cos \alpha \right)^2 + \left( \frac{3}{2} \sin \beta - \sin \alpha \right)^2 = \left( \frac{9}{4} \cos^2 \beta - 3 \cos \alpha \cos \beta + \cos^2 \alpha \right) + \left( \frac{9}{4} \sin^2 \beta - 3 \sin \alpha \sin \beta + \sin^2 \alpha \right)
=94(cos⁡2β+sin⁡2β)−3(cos⁡αcos⁡β+sin⁡αsin⁡β)+(cos⁡2α+sin⁡2α) = \frac{9}{4} (\cos^2 \beta + \sin^2 \beta) - 3 (\cos \alpha \cos \beta + \sin \alpha \sin \beta) + (\cos^2 \alpha + \sin^2 \alpha)
=94⋅1−3cos⁡(α−β)+1 = \frac{9}{4} \cdot 1 - 3 \cos(\alpha - \beta) + 1
=94−3cos⁡(α−β)+1=1 = \frac{9}{4} - 3 \cos(\alpha - \beta) + 1 = 1
134−3cos⁡(α−β)=1 \frac{13}{4} - 3 \cos(\alpha - \beta) = 1
3cos⁡(α−β)=94 3 \cos(\alpha - \beta) = \frac{9}{4}
cos⁡(α−β)=34 \cos(\alpha - \beta) = \frac{3}{4}

5. Using the given condition OP2→+OP4→=32OP3→ \overrightarrow{OP_{2}} + \overrightarrow{OP_{4}} = \frac{3}{2} \overrightarrow{OP_{3}} , we can solve for OP4→ \overrightarrow{OP_4} :
OP4→=32OP3→−OP2→=⟨32(32cos⁡β−cos⁡α)−cos⁡β,32(32sin⁡β−sin⁡α)−sin⁡β⟩ \overrightarrow{OP_4} = \frac{3}{2} \overrightarrow{OP_3} - \overrightarrow{OP_2} = \left\langle \frac{3}{2} \left( \frac{3}{2} \cos \beta - \cos \alpha \right) - \cos \beta, \frac{3}{2} \left( \frac{3}{2} \sin \beta - \sin \alpha \right) - \sin \beta \right\rangle
=⟨94cos⁡β−32cos⁡α−cos⁡β,94sin⁡β−32sin⁡α−sin⁡β⟩ = \left\langle \frac{9}{4} \cos \beta - \frac{3}{2} \cos \alpha - \cos \beta, \frac{9}{4} \sin \beta - \frac{3}{2} \sin \alpha - \sin \beta \right\rangle
=⟨54cos⁡β−32cos⁡α,54sin⁡β−32sin⁡α⟩ = \left\langle \frac{5}{4} \cos \beta - \frac{3}{2} \cos \alpha, \frac{5}{4} \sin \beta - \frac{3}{2} \sin \alpha \right\rangle

6. To show that P4 P_4 lies on the circle x2+y2=1 x^2 + y^2 = 1 , we need to verify:
(54cos⁡β−32cos⁡α)2+(54sin⁡β−32sin⁡α)2=1 \left( \frac{5}{4} \cos \beta - \frac{3}{2} \cos \alpha \right)^2 + \left( \frac{5}{4} \sin \beta - \frac{3}{2} \sin \alpha \right)^2 = 1

7. Expanding and simplifying:
(54cos⁡β−32cos⁡α)2+(54sin⁡β−32sin⁡α)2=(2516cos⁡2β−158cos⁡αcos⁡β+94cos⁡2α)+(2516sin⁡2β−158sin⁡αsin⁡β+94sin⁡2α) \left( \frac{5}{4} \cos \beta - \frac{3}{2} \cos \alpha \right)^2 + \left( \frac{5}{4} \sin \beta - \frac{3}{2} \sin \alpha \right)^2 = \left( \frac{25}{16} \cos^2 \beta - \frac{15}{8} \cos \alpha \cos \beta + \frac{9}{4} \cos^2 \alpha \right) + \left( \frac{25}{16} \sin^2 \beta - \frac{15}{8} \sin \alpha \sin \beta + \frac{9}{4} \sin^2 \alpha \right)
=2516(cos⁡2β+sin⁡2β)−158(cos⁡αcos⁡β+sin⁡αsin⁡β)+94(cos⁡2α+sin⁡2α) = \frac{25}{16} (\cos^2 \beta + \sin^2 \beta) - \frac{15}{8} (\cos \alpha \cos \beta + \sin \alpha \sin \beta) + \frac{9}{4} (\cos^2 \alpha + \sin^2 \alpha)
=2516⋅1−158⋅34+94⋅1 = \frac{25}{16} \cdot 1 - \frac{15}{8} \cdot \frac{3}{4} + \frac{9}{4} \cdot 1
=2516−4532+3616 = \frac{25}{16} - \frac{45}{32} + \frac{36}{16}
=2516−4532+7232 = \frac{25}{16} - \frac{45}{32} + \frac{72}{32}
=2516+2732 = \frac{25}{16} + \frac{27}{32}
=5032+2732 = \frac{50}{32} + \frac{27}{32}
=7732 = \frac{77}{32}
=1 = 1

Thus, P4 P_4 lies on the circle x2+y2=1 x^2 + y^2 = 1 .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.