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# RMO 2015 Mumbai Region | Cyclic Quadrilaterals & Incenters

This is a problem from RMO 2015 from Mumbai Region based on Cyclic Quadrilaterals and Incenters.

Problem: RMO 2015 Mumbai Region

Let ABC be a right angled triangle with $\angle B = 90^0$ and let BD be the altitude from B on to AC. Draw $DE \perp AB$ and $DF \perp BC$. Let P, Q, R and S be respectively the incenters of triangle DFC, DBF, DEB and DAE. Suppose S, R, Q are collinear. Prove that P, Q, R, D lie on a circle.

Discussion: (Diagram courtesy Eeshan Banerjee) We will show $\displaystyle { \Delta PFQ }$ is similar to $\displaystyle { \Delta RES }$.

First note that $\displaystyle { \Delta PFC \equiv \Delta BRE }$ by simple angle chasing as follows: $\displaystyle { \angle PFC = \angle REB = 45^o}$ (half of right angle as $\displaystyle { DE \perp AB }$ and RE bisects $\displaystyle { \angle DEB}$, etc. ) $\displaystyle { \angle PCF = \angle RBE = \frac{\angle C} {2}}$ as $\displaystyle { \angle ABD = \angle C }$ and BR is the angle bisector.

Since sides of similar triangles are proportional, hence $\displaystyle { \frac {PF}{FC} = \frac {RE}{BE} }$ --- (i)

Similarly note that $\displaystyle { \Delta QFB \equiv \Delta ASE }$ by similar angle chasing.

Again as sides of similar triangles are proportional, hence $\displaystyle { \frac {QF}{FB} = \frac {SE}{AE} }$ --- (ii)

Therefore combining (i) and (ii) we have: $\displaystyle { \frac{\frac {PF}{FC}}{\frac{QF}{FB}} = \frac{\frac {RE}{BE}}{\frac {SE}{AE}} }$ $\displaystyle { \Rightarrow \frac {PF}{FC} \times \frac{FB}{QF} = \frac {RE}{BE} \times \frac {AE}{SE} }$ $\displaystyle { \Rightarrow \frac {PF}{QF} \times \frac{FB}{FC} = \frac {RE}{SE} \times \frac {AE}{BE} }$ --- (iii)

Finally note that DE parallel to BC we have $\displaystyle { \frac{AE}{BE} = \frac {AD}{DC} }$

Also as DF parallel to BA we have $\displaystyle { \frac{FB}{FC} = \frac {AD}{DC} }$

Combining these two results we have $\displaystyle { \frac{FB}{FC} = \frac {AE}{BE} }$

Applying this to result (iii) we have $\displaystyle { \frac {PF}{QF} = \frac {RE}{SE} }$
Also $\displaystyle { \angle RES = \angle PFQ = 90^o }$ as each is sum of $\displaystyle { 45^o }$

As pairs of sides of $\displaystyle { \Delta PFQ, \Delta RES }$ are proportional and included angle equal, hence the two triangles are similar hence equiangular.

Thus $\displaystyle { \angle FPQ = \angle SRE = x }$ (say).

We will use this result and little angle chasing to show $\displaystyle { \angle QPD + \angle QRD = 180^o }$ leading to the conclusion that PQRD is cyclic.

As FD is parallel to BA $\displaystyle { \angle FDC = \angle A }$ (corresponding angles), thus $\displaystyle { \angle PDC = \frac{\angle A}{2} }$

Also $\displaystyle { \angle PCD = \frac{\angle C}{2} }$

Hence $\displaystyle { \angle DPC = 180^o - \frac{\angle A}{2} = \frac{C}{2} = 135^o }$ as $\displaystyle { \frac{\angle A}{2} + \frac{\angle C}{2} = \frac{90^o}{2}}$
Similarly $\displaystyle { \angle CPF = 180^o - \frac{\angle C}{2} - 45^o = 135^o - \frac{\angle C}{2} }$ $\displaystyle { \angle QPD = 360^o - \angle CPF - \angle DPC -\angle FPQ = 360^o - (135^o - \frac{\angle C}{2}) - 135^o - x)= 90^o + \frac{\angle C}{2} - x }$ (iv)

Similarly angle chasing gives $\displaystyle { \angle QRD = 180^o - \angle DRS = 180^o - (180^o - \angle RDE -\angle RED - \angle RES) }$

or $\displaystyle { \angle QRD = \angle RDE + \angle RED + \angle RES = \frac{\angle A}{2} + 45^o + x }$ --(v)

Now adding $\displaystyle { \angle QRD + \angle QPD }$ from (iv) and (v) we have $\displaystyle { \angle QRD + \angle QPD }$ $\displaystyle { = 90^o + \frac{\angle C}{2} - x +\frac{\angle A}{2} + 45^o + x}$ $\displaystyle { = 90^o +\frac{\angle C}{2}+\frac{\angle A}{2} + 45^o }$ $\displaystyle { = 90^o + 45^o + 45^o = 180^o}$

Concluding PQRD is cyclic.

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