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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] /* AMC 10A 2025 Problem — Equilateral Triangle Trisectors Diagram */ import olympiad; size(220); pair A = (0,0); pair B = (10,0); pair C = (5,8.660254037844386); // height = 5*sqrt(3) // Triangle draw(A--B--C--cycle, linewidth(1)); // Mark vertices dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); // Length of trisectors (visual only) real L = 8.8; // Trisectors from each vertex (20° and 40° offsets from sides) draw(A--(A + L*dir(20)), blue+linewidth(0.8)); draw(A--(A + L*dir(40)), blue+linewidth(0.8)); draw(B--(B + L*dir(140)), blue+linewidth(0.8)); draw(B--(B + L*dir(160)), blue+linewidth(0.8)); draw(C--(C + L*dir(260)), blue+linewidth(0.8)); draw(C--(C + L*dir(280)), blue+linewidth(0.8)); // Intersection points (precomputed numerically) 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); // Hexagon interior filldraw(P1--P2--P3--P4--P5--P6--cycle, lightred+opacity(0.4), red+linewidth(0.8)); // Mark intersection region label("$P_1$", P1, NW); label("$P_2$", P2, N); label("$P_3$", P3, NE); label("$P_4$", P4, SE); label("$P_5$", P5, S); label("$P_6$", P6, SW); label("Hexagon formed by the middle 20° trisectors", (5, -1.2)); [/asy]

~Avs2010

~text edits by i_am_not_suk_at_math (saharshdevaraju 22:31, 7 November 2025 (EST)saharshdevaraju) also, please stop trying to remove my contributions to this whoever is doing so.

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 [SIMPLE: ONLY ISOSCELES TRIANGLES]

https://imgur.com/a/Hm7Bybf

Angle A is split into three so the triangle $AEB$ is an isosceles triangle because the bottom angles 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 angles $C$ 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), we have: \[x + x = 240 \implies x = 120^\circ.\]

Thus, angle $G$ is $100^\circ$ because of triangle $FGE$.

Now find H. In isosceles triangle $AHB$, angles $A$ and $B$ are $40^\circ$, so angle $H$ is $100^\circ$.

Now find I. The red hexagon’s interior angles sum to $720^\circ$, and angle $I$ is congruent to the angle across from it by symmetry.

Let $I$ and its symmetric angle be $x$. Then: \[2x + 140(E) + 2(100)(G\ \&\ \text{its symmetry}) + 100(H) = 720\] \[\implies x = 140^\circ.\]

The smallest angle is $100^\circ$.

~PUER_137

Solution 5 [Most outer triangle]

Using the outside triangle made by a trisection, we know that two of the angles are $20^\circ$ and $60^\circ,$ it follows that the third angle in the triangle, the foot of a trisection is $100^\circ.$

We then take a different triangle, that utilizes two of the same lines as the first triangle we examined and also has the $100^\circ$ angle. This time, we can use the $40^\circ$ angle made by two of the trisections, and we get a triangle with angles $40^\circ, 40^\circ, 100^\circ.$

We can look at a dart-like figure (inverted kite) and we get by symmetry, the angle opposite of the initial $40^\circ$ angle is also $40^\circ,$ there is also the middle angle formed by the trisection, $20^\circ.$ Using the dart theorem (I don't know why this isn't a thing when I search it up) we find that one angle in the hexagon is $100^\circ$ and by symmetry, that is the smallest angle, so the answer is $\boxed{\textbf{(C) }100}$

~happyfish0922

Chinese Video Solution

https://www.bilibili.com/video/BV1SV2uBtESe/

~metrixgo

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

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

These problems are copyrighted © by the Mathematical Association of America.

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