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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Properties of Triangles

Olympiad workbook · IOQM · Properties of Triangles

The incircle of a scalene triangle A B C touches B C at D, C A at E...

The incircle Γ\Gamma of a scalene triangle ABCA B C touches BCB C at D,CAD, C A at EE and ABA B at FF. Let rAr_A be the radius of the circle inside ABCA B C which is tangent to Γ\Gamma and the sides ABA B and ACA C. Define rBr_B and rCr_C similarly. If rA=16,rB=25r_A=16, r_B=25 and rC=36r_C=36, determine the radius of Γ\Gamma.The incircle of a scalene triangle A B C touches B C at D, C A at E and A B at F . Let r_A be the radius of the circle

Correct answer

74

Step-by-step solution

Let inradius of △ABC\triangle A B C be rr The incircle of a scalene triangle A B C touches B C at D, C A at E and A B at F . Let r_A be the radius of the circle
 In △IDC:IC=rcosec⁡C2\text { In } \triangle I D C: I C=r \operatorname{cosec} \frac{C}{2} and I1C=36cosec⁡C2I_1 C=36 \operatorname{cosec} \frac{C}{2} ∴I1=rcosec⁡C2−36cosec⁡C2=36+r⇒r−36r+36=sin⁡C2\begin{aligned} & \therefore I_1=r \operatorname{cosec} \frac{C}{2}-36 \operatorname{cosec} \frac{C}{2}=36+r \\ & \Rightarrow \frac{r-36}{r+36}=\sin \frac{C}{2} \end{aligned} Similarly r−25r+25=sin⁡B2\frac{r-25}{r+25}=\sin \frac{B}{2} and r−16r+16=sin⁡A2\frac{r-16}{r+16}=\sin \frac{A}{2} ∵sin⁡A2=sin⁡π−(B+C)2∴sin⁡A2=cos⁡B2cos⁡C2−sin⁡B2sin⁡C2\begin{aligned} & \because \quad \sin \frac{A}{2}=\sin \frac{\pi-(B+C)}{2} \\ & \therefore \quad \sin \frac{A}{2}=\cos \frac{B}{2} \cos \frac{C}{2}-\sin \frac{B}{2} \sin \frac{C}{2} \end{aligned} ⇒r−16r+16=1−(r−25r+25)21−(r−36r+36)2−r−25r+25⋅r−36r+36⇒r−16r+16=10rr+25⋅12rr+36−(r−25)(r−36)(r+25)(r+36)⇒(r−16)(r+25)(r+36)=(r+16)(120r−(r2−61r+900))⇒r(r2+61r+900−181r+900+r2)=16(r2+61r+900+181r−900−r2)⇒2r2−120r+1800=16×242⇒r2−60r+900=1936⇒r2−60r−1036=0⇒(r+14)(r−74)=0∴r=74\begin{aligned} \Rightarrow & \frac{r-16}{r+16}=\sqrt{1-\left(\frac{r-25}{r+25}\right)^2} \sqrt{1-\left(\frac{r-36}{r+36}\right)^2} \\ & -\frac{r-25}{r+25} \cdot \frac{r-36}{r+36} \\ \Rightarrow & \frac{r-16}{r+16}=\frac{10 \sqrt{r}}{r+25} \cdot \frac{12 \sqrt{r}}{r+36}-\frac{(r-25)(r-36)}{(r+25)(r+36)} \\ \Rightarrow & (r-16)(r+25)(r+36)=(r+16)(120 r- \\ & \left.\left(r^2-61 r+900\right)\right) \\ \Rightarrow & r\left(r^2+61 r+900-181 r+900+r^2\right)=16\left(r^2+61 r\right. \\ & \left.+900+181 r-900-r^2\right) \\ \Rightarrow & 2 r^2-120 r+1800=16 \times 242 \\ \Rightarrow & r^2-60 r+900=1936 \\ \Rightarrow & r^2-60 r-1036=0 \\ \Rightarrow & (r+14)(r-74)=0 \\ \therefore & r=74\end{aligned}
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