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 4 of 5: Compare the two circles

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I\text{ lies inside }(BOC),\quad P\text{ lies outside }(BOC)

Detailed analysis

The points O and I are on the same side of BC, and angle BIC is larger than angle BOC; hence I is inside the circumcircle of the isosceles triangle BOC. Since O is the midpoint of IP, P is the reflection of I in O, so P is outside that circle and on the same side of BC as O, as in the official construction.