SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Question

Let ABC be an acute triangle with AT, AS respectively the internal and external angle bisectors of ABC and T, S BC. On the circle with diameter TS, take an arbitrary point P that lies inside the triangle ABC. Denote D, E, F, I as the incenter of triangle PBC, PCA, PAB, ABC. Prove that four lines AD, BE, CF and IP are concurrent. FIGURE_0

Accepted Answer

First, we note that the circle of diameter

T S

is the Apollonius circle of triangle

A B C

then

\frac{B P}{C P} = \frac{B A}{C A} = \frac{B T}{C T}

or

\frac{B P}{B A} = \frac{C P}{C A},

which implies that the bisector of angle

B

in triangle

A B P

and the bisector of angle

C

in triangle

A C P

pass through the same point on

A P

. Denote that point as

K

. So

C E

,

B F

,

A P

are concurrent at

K

.

It is easy to see that

I \in A T

. Consider triangle

A P T

and we have

\frac{I A}{I T} \cdot \frac{K P}{K A} \cdot \frac{D T}{D P} = \frac{B A}{B T} \cdot \frac{B P}{B A} \cdot \frac{B T}{B P} = 1

then by applying Ceva's theorem, we can see that three lines

A D

,

I P

,

K T

are concurrent.

Continue, consider triangle

K B C

and we have

\frac{F K}{F B} \cdot \frac{E C}{E K} = \frac{P K}{P B} \cdot \frac{P C}{P K} = \frac{P C}{P B} = \frac{T C}{T B}

or

\frac{F K}{F B} \cdot \frac{E C}{E K} \cdot \frac{T B}{T C} = 1 (2)

then

K T

,

B E

,

C F

are also concurrent.

Finally, suppose that

B E

,

C F

,

A D

are concurrent at

X

then denote

B E \cap C F = X

, we will have

X \in K T

based on (2), but

X \in A D

then

X \in K T \cap A D

. Therefore, based on (1), we have

X \in I P

, then the result will follow.