Indian National Mathematic Olympiad 2016 Q1

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Problem


Let ABC be triangle in which AB = AC. Suppose the orthocentre of the triangle lies on the incircle. Find the ratio AB/BC

Solution

Method 1

Proof:

Given AB=AC, AD is perpendicular bisector of BC and angle bisector of {\angle BAC}

So, incenter and orthocenter lies on AD

Since H (Orthocenter) lies on incircle

\displaystyle HI = ID \ \ \ \ \ (1)

But {BE \perp AC} and {GF \perp AC} ,

\displaystyle \implies BE \parallel GF

Since, {GI \parallel BH}

\displaystyle \implies \frac{DG}{BG} = \frac{DI}{IH} = 1

\displaystyle \implies BG = DG

Since D bisects BC, so {BD = DC}

Let’s say BG = x,

So, GD =x; BD = 2x; DC = 2x; BC = 4x; GC = 3x


CF and CD are tangent to incircle, so

\displaystyle CD = CF = 2x


Now, {\angle ACD = \angle FCG; \angle ADC = \angle CFG; \angle DAC = \angle FGC}

\displaystyle \implies \Delta GFC \sim \Delta ADC

\displaystyle \implies \frac{AC}{DC} = \frac{GC}{FC} = \frac{3x}{2x} = \frac{3}{2}

So,

\displaystyle \frac{AB}{BC} = \frac{AC}{2*DC}

\displaystyle \implies \frac{AB}{BC} = \frac{3}{4}

\Box

Method 2

Proof:

Given AB=AC, AD is perpendicular bisector of BC and angle bisector of {\angle BAC}

So, incenter and orthocenter lies on AD

Since H (Orthocenter) lies on incircle

\displaystyle HI = ID \ \ \ \ \ (2)

But {BG \perp AC } and {IF \perp AC} and {DE \perp AC},

\displaystyle \implies BG \parallel IF \parallel DE

\displaystyle \implies \frac{GF}{FE} = \frac{HI}{ID} = 1

\displaystyle \implies GF = FE

Since {DE \parallel BG}

\displaystyle \implies \frac{GE}{EC} = \frac{BD}{DC} = 1

\displaystyle \implies BD = DC

Let’s say GF = a,

So, FE =a; GE = 2a; EC = 2a; GC = 4a; FC = 3a


CF and CD are tangent to incircle, so

\displaystyle CD = CF = 3a


Now, {\angle ACD = \angle ECD; \angle ADC = \angle DEC; \angle DAC = \angle EDC}

\displaystyle \implies \Delta DEC \sim \Delta ADC

\displaystyle \implies \frac{AC}{DC} = \frac{DC}{EC} = \frac{3a}{2a} = \frac{3}{2}

So,

\displaystyle \frac{AB}{BC} = \frac{AC}{2*DC}

\displaystyle \implies \frac{AB}{BC} = \frac{3}{4}

\Box

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