Exam

Olympiad workbook

Subject

IOQM

Chapter

Properties of Triangles

Olympiad workbook · IOQM · Properties of Triangles

Let A B C be a triangle with A B=A C . Let D be a point on the...

Let $A B C$ be a triangle with $A B=A C$. Let $D$ be a point on the segment $B C$ such that $B D=48 \frac{1}{61}$ and $D C=61$. Let $E$ be a point on $A D$ such that $C E$ is perpendicular to $A D$ and $D E=11$. Find $A E$.

Step-by-step solution

Let $B D=48 \frac{1}{61}=\alpha, M$ is mid-point of $B C$. $$\begin{aligned} & C D=61=\beta \\ & B M=C M=\frac{\beta+\alpha}{2} \\ & D M=B M-B D=\frac{\beta-\alpha}{2} \end{aligned}$$ Let $A E=x, A B=A C=y, A M=h, C E=z$ In $\triangle C D E$, $$\begin{aligned} & C E^2=C D^2-D E^2=61^2-11^2=50 \times 72 \\ & C E=60=z ...(i)\\ & \text { In } \triangle A C E, y^2=x^2+60^2 \\ & \quad y^2=x^2+3600 ...(ii) \end{aligned}$$ In $\triangle A C M, y^2=h^2+\left(\frac{\beta+\alpha}{2}\right)^2$...(iii) In $\triangle A D M,(x+11)^2=h^2+\left(\frac{\beta-\alpha}{2}\right)^2$...(iv) (ii) \& (iii) $\Rightarrow x^2+3600=h^2+\left(\frac{\beta+\alpha}{2}\right)^2$...(v) $$\begin{aligned} & \text { (iv) } &(\mathrm{v}) \Rightarrow(x+11)^2-x^2-3600=\left(\frac{\beta-\alpha}{2}\right)^2-\left(\frac{\beta+\alpha}{2}\right)^2 \\ & 22 x+121-3600=-\alpha \beta \\ & =-\frac{(48 \times 61)+1}{61} \times 61 \\ & 22 x=3600-121-2929 \\ & x=25=A E\end{aligned}$$