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2025 AMC 10A Problems/Problem 6

In an equilateral triangle each interior angle is trisected by a pair of rays. The intersection of the interiors of the middle 20°-angle at each vertex is the interior of a convex hexagon. What is the degree measure of the smallest angle of this hexagon?

$\textbf{(A) } 80 \qquad\textbf{(B) } 90 \qquad\textbf{(C) } 100 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

Diagram

[asy] import olympiad; size(250); // Equilateral triangle vertices pair A = (0,0); pair B = (10,0); pair C = (5,8.660254037844386); // height = 5*sqrt(3) // Draw triangle draw(A--B--C--cycle, linewidth(1)); // Draw trisectors (hardcoded 20° from sides, ending inside triangle) real trisectorLength = 8.7939; // length of rays inside triangle draw(A--(A + trisectorLength*dir(20)),blue); draw(A--(A + trisectorLength*dir(40)),blue); draw(B--(B + trisectorLength*dir(140)),blue); draw(B--(B + trisectorLength*dir(160)),blue); draw(C--(C + trisectorLength*dir(260)),blue); draw(C--(C + trisectorLength*dir(280)),blue); pair P1=(4.07604, 3.4202); pair P2=(5, 4.1955); pair P3=(5.92396, 3.4202); pair P4=(6.13341, 2.23238); pair P5=(5, 1.81985); pair P6=(3.86659, 2.23238); filldraw(P1--P2--P3--P4--P5--P6--cycle, lightred); [/asy] ~Avs2010

Video Solution

https://youtu.be/l1RY_C20Q2M

Solution 1

Assume you have a diagram in front of you.

Because each angle of the triangle is trisected, we have 9 $20^\circ$ angles. Using a side of the triangle as a base, we have an isosceles triangle with two $20^\circ$ angles. Using this we can show that the third angle is $140^\circ$.

Following that, we use the vertex angles to show that one angle of the hexagon is $140^\circ$. And with rotational symmetry, three. The average of all 6 angles has to be $120^\circ$, so the answer is $\boxed{\textbf{(C) }100}$ - SpectralScholar

Solution 2

It is obvious that of the 6 angles inside the convex hexagon, there are only two different angle measures, 3 of one and 3 of another. A convex quadrilateral formed by the 2 rays of any angle in the equilateral triangle and two sides of the convex hexagon will have a total degree of 360.

Therefore, we have: $3a+3b=720 \implies a+b=240$ (total sum of all angles in a convex hexagon is 720) and also $20+2a+b=360 \implies 2a+b=340$ (the rays will form an inner angle of $\frac{60}{3}=20$ degrees). Subtracting the two equations yields $a=100$ and $b=140$. Hence our smallest angle in this convex hexagon is $\boxed{\textbf{(C) }100}$. ~hxve

Solution 3 (cheese)

Notice that only answer choices $(a)$ and $(c)$ sum to 180, a familiar number, and since $(a)$ is not a common answer, choose $(c)$ Note: this is a super informal way to do this, use only if you can't draw a picture or have no idea. 17:51, 6 November 2025 (EST)~Pungent_Muskrat

Solution 4 [NO ALGEBRA: PURE ANGLE CHASING]

[img]https://imgur.com/a/Hm7Bybf[/img] Angle A is split into three so the triangle $AEB$ is an isosceles triangle because the bottom angle A and B are congruent and both $20^\circ$. Therefore angle E is $140^\circ$ and the vertical angle in the hexagon is also $140^\circ$. Now find G. Triangle $CJD$ is isosceles with angle $C$ and $D$ being $80^\circ$ because angle J in that triangle is $20^\circ$. Now angles $C$, $D$, and $E$ are known and sum to $80+80+140=300$. the pentagon $DCFE$ and its other vertex (not named in my image) sum to $540^\circ$. So subtracting angles $C$, $D$, $E$, and knowing that $F$ (let it be x) is congruent (due to symmetry) to the other vertex angle (not named in my image) $x+x=240 \implies x=120^\circ$. Thus angle $G$ is $100^\circ$ because of triangle $FGE$. Now find H. In isosceles triangle $AHB$ $A$ and $B$ are $40^\circ$ so angle $H$ is $100^\circ$. Now find I. The Red Hexagon degrees sum to $720^\circ$ and angle I is congruent to the angle across from it by symmetry. Let and I and its symmetric angle be x. $2x+140(E)+2*100(G & its symmetry)+100(H)=720 \implies x=140$ (Error compiling LaTeX. Unknown error_msg)

The smallest angle is $100^\circ$

~PUER_137

Video Solution (Done in 1 Min)

https://youtu.be/qVm7neHfDrI?si=n7nLnWY_p1SLXoxr ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution by Daily Dose of Math

https://youtu.be/gPh9w3X3QSw

~Thesmartgreekmathdude

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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