Olympiad workbook
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Subject IOQM
Chapter Vector Algebra Olympiad workbook · IOQM · Vector Algebra
In a parallelogram A B C D , the point P on a segment A B is taken...
In a parallelogram
A B C D A B C D A BC D , the point
P P P on a segment
A B A B A B is taken such that
A P A B = 61 2022 \frac{A P}{A B}=\frac{61}{2022} A B A P = 2022 61 and a point
Q Q Q on the segment
A D A D A D is taken such that
A Q A D = 61 2065 \frac{A Q}{A D}=\frac{61}{2065} A D A Q = 2065 61 . If
P Q P Q PQ intersects
A C A C A C at
T T T , find
A C A T \frac{A C}{A T} A T A C to the nearest integer.
Step-by-step solution
Here,
α = 61 2022 \alpha=\frac{61}{2022} α = 2022 61 and
β = 61 2065 \beta=\frac{61}{2065} β = 2065 61 t ⃗ = b ⃗ + d ⃗ 1 + λ = α μ b ⃗ + β d ⃗ 1 + μ \vec{t}=\frac{\vec{b}+\vec{d}}{1+\lambda}=\frac{\alpha \mu \vec{b}+\beta \vec{d}}{1+\mu} t = 1 + λ b + d = 1 + μ αμ b + β d
So,
1 1 + λ = α μ 1 + μ \frac{1}{1+\lambda}=\frac{\alpha \mu}{1+\mu} 1 + λ 1 = 1 + μ αμ and
1 1 + λ = β 1 + μ \frac{1}{1+\lambda}=\frac{\beta}{1+\mu} 1 + λ 1 = 1 + μ β ⇒ α μ = β hence 1 1 + λ = β 1 + β α ⇒ 1 + λ = α + β α β \begin{aligned}
& \Rightarrow \quad \alpha \mu=\beta \text { hence } \frac{1}{1+\lambda}=\frac{\beta}{1+\frac{\beta}{\alpha}} \\
& \Rightarrow \quad 1+\lambda=\frac{\alpha+\beta}{\alpha \beta}
\end{aligned} ⇒ αμ = β hence 1 + λ 1 = 1 + α β β ⇒ 1 + λ = α β α + β
Clearly,
A C A T = 1 + λ 1 = α + β α β = 1 α + 1 β = 2022 61 + 2065 61 = 4087 61 = 67 \begin{aligned} \frac{A C}{A T}=\frac{1+\lambda}{1} & =\frac{\alpha+\beta}{\alpha \beta}=\frac{1}{\alpha}+\frac{1}{\beta} \\ & =\frac{2022}{61}+\frac{2065}{61} \\ & =\frac{4087}{61}=67\end{aligned} A T A C = 1 1 + λ = α β α + β = α 1 + β 1 = 61 2022 + 61 2065 = 61 4087 = 67
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