2025 AIME II Problem 5

Question

Suppose

\triangle ABC

has angles

\angle BAC = 84^\circ,

\angle ABC = 60^\circ,

and

\angle ACB = 36^\circ.

Let

D,

E,

and

F

be the midpoints of sides

\overline{BC},

\overline{AC},

and

\overline{AB},

respectively. The circumcircle of

\triangle DEF

intersects

\overline{BD},

\overline{AE},

and

\overline{AF}

at points

G,

H,

and

J,

respectively. The points

G,

D,

E,

H,

J,

and

F

divide the circumcircle of

\triangle DEF

into six minor arcs, as shown. Find

\overset{\frown}{DE} + 2 \cdot \overset{\frown}{HJ} + 3 \cdot \overset{\frown}{FG},

where the arcs are measured in degrees.

Answer

Answer: 336

Concepts:inscribed angle angle chasing isosceles triangle

Difficulty rating: 2720

Small hint:

The circle through the midpoints is the nine-point circle, so

G,

H,

J

are the feet of the altitudes, and

\triangle DEF

has the same angles as

\triangle ABC

Big hint:

Since

\angle BJC = \angle BHC = 90^\circ,

D

is equidistant from

B,

C,

H,

J;

isosceles triangles turn the arcs into angles of

\triangle ABC

Solution:

The medial triangle

DEF

has sides parallel to those of

ABC,

so

\angle FDE = 84^\circ,

\angle DEF = 60^\circ,

and

\angle DFE = 36^\circ.

Its circumcircle is the nine-point circle, whose second intersections with the sides of

ABC

are the feet of the altitudes:

G

is the foot from

A,

H

the foot from

B,

and

J

the foot from

C.

By the inscribed angle theorem,

\overset{\frown}{DE} = 2\angle DFE = 72^\circ.

For

\overset{\frown}{FG}:

since

\overline{DF} \parallel \overline{CA}

and

G

lies on ray

DB,

the angle

\angle FDG

equals the angle between lines

CA

and

CB,

which is

36^\circ,

so

\overset{\frown}{FG} = 2 \cdot 36^\circ = 72^\circ.

For

\overset{\frown}{HJ}:

because

\angle BJC = \angle BHC = 90^\circ,

both

H

and

J

lie on the circle with diameter

\overline{BC}

centered at

D,

so

DJ = DB

and

DH = DC.

Isosceles triangle

BDJ

gives

\angle JDB = 180^\circ - 2 \cdot 60^\circ = 60^\circ,

and isosceles triangle

CDH