USAJMO 2026 (Problem 3)

Let

ABC

be an acute scalene triangle with no angle equal to

60^\circ

. Let

\omega

be the circumcircle of

ABC

. Let

\Delta_B

be the equilateral triangle with three vertices on

\omega

, one of which is

B

. Let

\ell_B

be the line through the other two vertices of

\Delta_B

. Define

\Delta_C

and

\ell_C

analogously. Let

Y=AC\cap\ell_B

and

Z=AB\cap\ell_C

. Suppose the circumcircle of

AYZ

intersects

\omega

again at

P\ne A

. Let

D=BC\cap YZ

and

E=PA\cap YZ

. Prove that

PE=PD

Prove that

\boxed{p=\frac{ab+bc+ca}{a+b+c}}.

By Hints 9 and 11,

y=\frac{b(ab+bc+ca)}{b^2-ac},

z=\frac{c(ab+bc+ca)}{c^2-ab},

p=\frac{cz-by}{c+z-y-b}.

The numerator is

cz-by=\frac{c^2(ab+bc+ca)}{c^2-ab}-\frac{b^2(ab+bc+ca)}{b^2-ac}

=\frac{(ab+bc+ca)[c^2(b^2-ac)-b^2(c^2-ab)]}{(c^2-ab)(b^2-ac)}

=\frac{(ab+bc+ca)a(b^3-c^3)}{(c^2-ab)(b^2-ac)}

=\frac{a(b-c)(b^2+bc+c^2)(ab+bc+ca)}{(c^2-ab)(b^2-ac)}.

The denominator of

p

c+z-y-b=c-b+\frac{c(ab+bc+ca)}{c^2-ab}-\frac{b(ab+bc+ca)}{b^2-ac}

=\frac{(c-b)(c^2-ab)(b^2-ac)+(ab+bc+ca)[c(b^2-ac)-b(c^2-ab)]}{(c^2-ab)(b^2-ac)}

=\frac{(b-c)[(ab+bc+ca)^2-(c^2-ab)(b^2-ac)]}{(c^2-ab)(b^2-ac)}.

Now expand the expression in brackets:

(ab+bc+ca)^2-(c^2-ab)(b^2-ac)

=a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc

\quad-b^2c^2+ac^3+ab^3-a^2bc

=a[ab^2+ac^2+abc+2b^2c+2bc^2+b^3+c^3]

=a(a+b+c)(b^2+bc+c^2).

Therefore,

c+z-y-b=\frac{a(b-c)(a+b+c)(b^2+bc+c^2)}{(c^2-ab)(b^2-ac)}.

Hence,

p=\frac{\frac{a(b-c)(b^2+bc+c^2)(ab+bc+ca)}{(c^2-ab)(b^2-ac)}}{\frac{a(b-c)(a+b+c)(b^2+bc+c^2)}{(c^2-ab)(b^2-ac)}}

\boxed{p=\frac{ab+bc+ca}{a+b+c}}.