Download USAMTS Round 3 - Art of Problem Solving

Round: 3
Problem: 2/3/26
2/3/26
Let A1A2A3A4A5 be a regular
pentagon with side length 1. The
sides of the pentagon are extended to
form the 10-sided polygon shown in
bold at right. Find the ratio of the
area of quadrilateral A2A5B2B5
(shaded in the picture to the right) to
the area of the entire 10-sided
polygon.
Name: Jesse Doan
Username: jdoan17
ID Number: 31025
First, divide up 10 sided polygon by drawing all of the diagonals of
pentagon A1A2A3A4A5 .
Next, let’s give the tiny pentagon in pentagon A1A2A3A4A5 an area
of A. Secondly, the 5 isosceles triangles, with a shared side to
pentagon with area A, will be each given an area of B. And lastly, the 5
isosceles triangles who share two sides with the triangles with area B
will be given an area of C. Therefore, the new picture would be:
C
B
B
C
C
A
B
B
C
B
C
Round: 3
Problem: 2/3/26
2/3/26
Let A1A2A3A4A5 be a regular
pentagon with side length 1. The
sides of the pentagon are extended to
form the 10-sided polygon shown in
bold at right. Find the ratio of the
area of quadrilateral A2A5B2B5
(shaded in the picture to the right) to
the area of the entire 10-sided
polygon.
Name: Jesse Doan
Username: jdoan17
ID Number: 31025
Next, we will prove that triangle A3A4B1 is congruent to A3A4A1 .
Since they share the sides A3A4, we just need to prove that angle
A1A3A4 is congruent to angle B1A3A4 and then use ASA (AngleSide-Angle) to prove that the two triangles are congruent.
C
B
B
C
C
A
B
B
C
B
C
Since triangle A3A2A1 is isosceles and angle A3A2A1 is 108 degrees
by being an angle of a regular pentagon, angles A2A1A3 and A2A3A1
are equal to (180 - 108) / 2 = 36 degrees. Since angle A4A3A2 is equal
to 108 degrees and angle A2A3A1 is equal to 36 degrees, angle
A4A3A1 is equal to 108 - 36 = 72 degrees.
Next, since congruent angles B1 B2 B3 B4 B5 add up to 180 degrees,
B1 must equal 36 degrees. Since triangle B1A3A4 is isosceles and B1
is 36 degrees, angle B1A3A4 is equal to (180 - 36) / 2 = 72 degrees.
Therefore triangle A3A4B1 is congruent to triangle A3A4A1 because
they share side A3A4 and congruent angles, making them congruent
isosceles triangles.
Name: Jesse Doan
Username: jdoan17
ID Number: 31025
Round: 3
Problem: 2/3/26
2/3/26
Let A1A2A3A4A5 be a regular
pentagon with side length 1. The
sides of the pentagon are extended to
form the 10-sided polygon shown in
bold at right. Find the ratio of the
area of quadrilateral A2A5B2B5
(shaded in the picture to the right) to
the area of the entire 10-sided
polygon.
Knowing they are congruent triangles, this means each of the outer
isosceles triangles have an area equal to A + 3B + C.
A + 3B
+C
A + 3B + C
B
C
B
C
B
C
3B
C
B
C
+
+
C
+
+
A
A + 3B + C
B
A
A
3B
C
Now that each section of the 10-sided polygon is given an area in terms
of A, B, and C, we can calculate the ratio of the area of quadrilateral
A2A5B2B5 to the area of the entire 10-sided polygon.
Therefore, the area of the quadrilateral is
(A + 3B + C) + (A + 4B + 3C) + (A + 3B + C) = 3A + 10B + 5C
And the area of the entire 10-sided polygon is
5(A + 3B + C) + (A + 5B + 5C) = 6A + 20B + 10C
Therefore the ratio of the area of quadrilateral A2A5B2B5 to the area
of the entire 10-sided polygon is (3A + 10B + 5C) / (6A + 20B + 10C)
= ½ or 1:2