What's the length of the segment?

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In the above diagram, the tangent at P P and the secant Q R \overleftrightarrow{QR} intersect at T T and P S \overline{PS} bisects Q P R \angle{QPR} and Q S = a \overline{QS} = a and P T = a + 1 \overline{PT} = a + 1 .

Find the value of a a for which R S = 1 \overline{RS} = 1 .


The answer is 1.6180339887498948.

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1 solution

Rocco Dalto
May 1, 2021

λ = 1 2 m ( P M R ^ ) \lambda = \dfrac{1}{2}m(\widehat{PMR}) by tangent chord theorem m P Q R = λ \implies m\angle{PQR} = \lambda by inscribed angle theorem

m P S Q = 180 ( θ + λ ) m P S R = θ + λ = m S P T \implies m\angle{PSQ} = 180 - (\theta + \lambda) \implies m\angle{PSR} = \theta + \lambda = m\angle{SPT} \implies

P S T \triangle{PST} is an isosceles triangle with S T = P T = a + 1 \overline{ST} = \overline{PT} = a + 1 \implies

( a + 1 ) 2 = ( 2 a + 1 ) y (a + 1)^2 = (2a + 1)y by tangent secant theorem y = ( a + 1 ) 2 2 a + 1 = R T \implies y = \dfrac{(a + 1)^2}{2a + 1} = \overline{RT}

R S = S T R T = a + 1 ( a + 1 ) 2 2 a + 1 = a 2 + a 2 a + 1 = 1 \implies \overline{RS} = \overline{ST} - \overline{RT} = a + 1 - \dfrac{(a + 1)^2}{2a + 1} = \dfrac{a^2 + a}{2a + 1} = 1 \implies

a 2 + a = 2 a + 1 a 2 a 1 = 0 a^2 + a = 2a + 1 \implies a^2 - a - 1 = 0 and dropping the negative root \implies

a = 1 + 5 2 1.6180339887498948 a = \dfrac{1 + \sqrt{5}}{2} \approx \boxed{1.6180339887498948} .

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