Problem 4

Let triangle

ABC

with incenter

I

satisfy

AB<AC<BC

. Let

X

be a point on line

BC

, different from

C

, such that the line through

X

parallel to

AC

is tangent to the incircle. Similarly, let

Y

be a point on line

BC

, different from

B

, such that the line through

Y

parallel to

AB

is tangent to the incircle. Line

AI

intersects the circumcircle of triangle

ABC

again at

P

. Let

K

and

L

be the midpoints of

AC

and

AB

, respectively. Prove that

\angle KIL+\angle YPX=180^\circ

Step 2 of 4: The rhombus symmetry produces two new tangent lines

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In plain words

Point reflection through the incenter maps the incircle to itself and turns each side of the triangle into the other tangent line parallel to that side, and since it sends A to T, that other tangent line must pass through T.

TX\parallel AC,\ TY\parallel AB\ \text{are tangent to the incircle}

Detailed analysis

Point reflection in $I$ preserves the incircle and sends the tangent $AC$ to the other tangent parallel to it through $T$ ; by the definition of $X$ , this is $TX$ . Similarly $TY$ is tangent and $TY\parallel AB$ . We also need the order on $BC$ , not merely collinearity. Put $a=BC$ , $b=CA$ , $c=AB$ , $s=(a+b+c)/2$ , and choose coordinates $B=(0,0)$ , $C=(a,0)$ , $A=(u,v)$ . The incenter is $I=(aA+bB+cC)/(2s)$ and $T=2I-A$ . Intersecting $T+\lambda(C-A)$ and $T+\mu(B-A)$ with $BC$ gives $BX=\frac{a(s-b)}s$ and $BY=\frac{ac}s$ . Since $0<s-b<c<s$ , we have $0<BX<BY<a$ , so $B,X,Y,C$ occur in this order. This also identifies the intended tangent on each parallel pair and will fix all ray orientations below.