Problem 1

Let

ABC

be a triangle with

\angle A<60^\circ

. Let

X

and

Y

be the points on the sides

AB

and

AC

, respectively, such that

CA+AX=CB+BX

and

BA+AY=BC+CY

. Let

P

be the point in the plane such that the lines

PX

and

PY

are perpendicular to

AB

and

AC

, respectively. Prove that

\angle BPC<120^\circ

Step 1 of 5: Locate X and Y

Previous step Next step

AD=AE=\dfrac{CA+AB-BC}{2},\quad AX=\dfrac{AB+BC-CA}{2}=BD

Detailed analysis

Let I be the incenter, and let D,E be its perpendicular feet on AB,AC. Assume without loss of generality that AC is the longest side, so X lies on AD. The given length equations give AX=(AB+BC−CA)/2 and AY=(BC+CA−AB)/2. Since AD=AE=(CA+AB−BC)/2, we have BD=AX and CE=AY.