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 $ s=2 $ and let BD be the altitude from B on to AC. Draw $ DE \perp AB $ and $ s=2$ and $ DF \perp BC $ and $ s=2 $. 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 } $ and $ s=2 $ is similar to $ \displaystyle { \Delta RES } $ and $ s=2 $.

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

Since sides of similar triangles are proportional, hence $ \displaystyle { \frac {PF}{FC} = \frac {RE}{BE} } $ and $ s=2 $ --- (i)

Similarly note that $ \displaystyle { \Delta QFB \equiv \Delta ASE } $ and $ s=2 $ by similar angle chasing.

Again as sides of similar triangles are proportional, hence $ \displaystyle { \frac {QF}{FB} = \frac {SE}{AE} } $ and $ s=2 $ --- (ii)

Therefore combining (i) and (ii) we have:

$ \displaystyle { \frac{\frac {PF}{FC}}{\frac{QF}{FB}} = \frac{\frac {RE}{BE}}{\frac {SE}{AE}} } $ and $ s=2 $

$ \displaystyle { \Rightarrow \frac {PF}{FC} \times \frac{FB}{QF} = \frac {RE}{BE} \times \frac {AE}{SE} } $ and $ s=2 $

$ \displaystyle { \Rightarrow \frac {PF}{QF} \times \frac{FB}{FC} = \frac {RE}{SE} \times \frac {AE}{BE} } $ and $ s=2 $ --- (iii)

Finally note that DE parallel to BC we have $ \displaystyle { \frac{AE}{BE} = \frac {AD}{DC} } $ and $ s=2 $

Also as DF parallel to BA we have $ \displaystyle { \frac{FB}{FC} = \frac {AD}{DC} } $ and $ s=2 $

Combining these two results we have $ \displaystyle { \frac{FB}{FC} = \frac {AE}{BE} } $ and $ s=2 $

Applying this to result (iii) we have
$ \displaystyle { \frac {PF}{QF} = \frac {RE}{SE} } $ and $ s=2 $
Also $ \displaystyle { \angle RES = \angle PFQ = 90^o } $ and $ s=2 $ as each is sum of $ \displaystyle { 45^o } $ and $ s=2 $

As pairs of sides of $ \displaystyle { \Delta PFQ, \Delta RES } $ and $ s=2 $ are proportional and included angle equal, hence the two triangles are similar hence equiangular.

Thus $ \displaystyle { \angle FPQ = \angle SRE = x } $ and $ s=2 $ (say).

We will use this result and little angle chasing to show

$ \displaystyle { \angle QPD + \angle QRD = 180^o } $ and $ s=2 $ leading to the conclusion that PQRD is cyclic.

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

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

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

Similarly angle chasing gives $ \displaystyle { \angle QRD = 180^o - \angle DRS = 180^o - (180^o - \angle RDE -\angle RED - \angle RES) } $ and $ s=2 $

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

Now adding $ \displaystyle { \angle QRD + \angle QPD } $ and $ s=2 $ from (iv) and (v) we have

$ \displaystyle { \angle QRD + \angle QPD }$ and $s=2$
$ \displaystyle { = 90^o + \frac{\angle C}{2} - x +\frac{\angle A}{2} + 45^o + x} $ and $ s=2 $
$ \displaystyle { = 90^o +\frac{\angle C}{2}+\frac{\angle A}{2} + 45^o } $ and $ s=2 $
$ \displaystyle { = 90^o + 45^o + 45^o = 180^o} $ and $ s=2 $

Concluding PQRD is cyclic.

Chatuspathi:

• Paper: RMO 2015 (Mumbai Region)
• What is this topic: Geometry
• What are some of the associated concepts: Similarity of triangles, cyclic quadrilateral, angle chasing
• Where can learn these topics: Cheenta I.S.I. & C.M.I. course,Cheenta Math Olympiad Program, discuss these topics in the ‘Geometry’ module.
• Book Suggestions: Challenges and Thrills of Pre-College Mathematics by Venkatchala