This is a question from the UKMT Maclaurin Olympiad, in which the solution is pretty straightforward as well! There are 2 (or more) ways of approaching this problem, but you could only rely on basic geometry!
Put your answer as a decimal, for example 0.4, 0.025 etc
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Let ∠ S P T = ∠ T P U = θ . Then ∠ U P Q = 9 0 ∘ − 2 θ , and
∣ U R ∣ = ∣ R Q ∣ − ∣ P Q ∣ tan ∠ U P Q = 2 − 2 tan ( 9 0 ∘ − 2 θ ) = 2 − 2 cot ( 2 θ ) = 2 − 2 × 2 tan θ 1 − tan 2 θ = 2 − 2 × 2 × 2 1 1 − 4 1 = 2 1 = 0 . 5 Note that tan θ = 2 1
Let ∠ S P T = ∠ T P U = α . Then tan α = 2 1 , tan ( 2 α ) = 2 − ∣ U R ∣ 2 = 1 − 4 1 2 × 2 1 = 3 4 ⟹ ∣ U R ∣ = 2 1 = 0 . 5 .
i couldnt find where to post a solution. i find a way without trigonometry. lets build perpendicular from T to PU and call it O. PTU and OTU are similar so OU/TU=TU/PR -> sqrt5 OU=TU. TU^2=TR^2+RU^2, which is (sqrt5 OU)^2=OU^2+1 -->OU=RU=0,5
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How do you know that △ P T U and △ O T U are similar?
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nice point! he needs to show that..
triangle PTO & triangle PTS are congruent by (ASA) congruency, so by (CPCT) ST = OT = 1. Again consider quad(PSTO) it's a cyclic quad... so (angle SPO) = (angle OTR) = 2 theta... similarly triangle OTU and triangle RTU are congruent by (RHS) congruency … which further gives (angle OTU) = (angle RTU) = theta ..by CPCT.
Now, (angle PTU) = angle PTO + angle RTU = (90 - theta) + theta = 90... which makes PTU and OTU similar triangles...
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