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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Properties of Triangles

Olympiad workbook · IOQM · Properties of Triangles

Let ABC be an isosceles triangle with AB=BC . A trisector of B...

Let ABC be an isosceles triangle with AB=BC\mathrm{AB}=\mathrm{BC}. A trisector of ∠B\angle \mathrm{B} meets AC at D . If AB,AC\mathrm{AB}, \mathrm{AC} and BDB D are integers and AB−BD=3A B-B D=3, find ACA C.Let ABC be an isosceles triangle with AB=BC . A trisector of B meets AC at D . If AB, AC and B D are integers and A B-B

Correct answer

26

Step-by-step solution

Let ABC be an isosceles triangle with AB=BC . A trisector of B meets AC at D . If AB, AC and B D are integers and A B-B
I1sin⁡(π2+θ)=I3sin⁡(π2−3θ)⇒ put I3=I1−3⇒I1=34cosec⁡2θ If cosec⁡2θ=36 i.e. cosec⁡2θ=6\begin{aligned} & \frac{I_1}{\sin \left(\frac{\pi}{2}+\theta\right)}=\frac{I_3}{\sin \left(\frac{\pi}{2}-3 \theta\right)} \\ & \Rightarrow \text { put } I_3=I_1-3 \\ & \Rightarrow I_1=\frac{3}{4} \operatorname{cosec}^2 \theta \\ & \text { If } \operatorname{cosec}^2 \theta=36 \\ & \text { i.e. } \operatorname{cosec}^2 \theta=6\end{aligned}  i.e. sin⁡θ=16I1=34×36=27I3=24I1sin⁡(π2−θ)=DCsin⁡4θI1sin⁡(π2+θ)=ADsin⁡2θ⇒I2=I1sin⁡4θcos⁡θ+I1sin⁡2θcos⁡θ⇒I2=I1×2×sin⁡3θ\begin{aligned} & \text { i.e. } \sin \theta=\frac{1}{6} \\ & I_1=\frac{3}{4} \times 36=27 \\ & I_3=24 \\ & \frac{I_1}{\sin \left(\frac{\pi}{2}-\theta\right)}=\frac{D C}{\sin 4 \theta} & \frac{I_1}{\sin \left(\frac{\pi}{2}+\theta\right)}=\frac{A D}{\sin 2 \theta} \\ & \Rightarrow I_2=\frac{I_1 \sin 4 \theta}{\cos \theta}+\frac{I_1 \sin 2 \theta}{\cos \theta} \\ & \Rightarrow I_2=I_1 \times 2 \times \sin 3 \theta\end{aligned} ⇒34cosec⁡2θ×2×sin⁡3θ⇒32(3sin⁡θ−4sin⁡θ)I2=32(3×6−46)=32(18−23)=32(54−23)=32×523I2=26\begin{aligned} \Rightarrow & \frac{3}{4} \operatorname{cosec}^2 \theta \times 2 \times \sin 3 \theta \\ \Rightarrow & \frac{3}{2}\left(\frac{3}{\sin \theta}-4 \sin \theta\right) \\ \mathrm{I}_2 & =\frac{3}{2}\left(3 \times 6-\frac{4}{6}\right) \\ & =\frac{3}{2}\left(18-\frac{2}{3}\right) \\ & =\frac{3}{2}\left(\frac{54-2}{3}\right) \\ & =\frac{3}{2} \times \frac{52}{3} \\ \mathrm{I}_2 & =26\end{aligned}
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