Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2025 AMC 10A Problems/Problem 6: Difference between revisions

← Older edit
Avs2010 (talk | contribs)
editor
40 edits
 
(27 intermediate revisions by 11 users not shown)
Line 1: Line 1:
==Problem==
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?
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?


<imath>\textbf{(A) } 80 \qquad\textbf{(B) } 90 \qquad\textbf{(C) } 100 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120</imath>
<imath>\textbf{(A) } 80 \qquad\textbf{(B) } 90 \qquad\textbf{(C) } 100 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120</imath>


===Diagram===
==Diagram==
<asy>
<asy>
/* AMC 10A 2025 Problem — Equilateral Triangle Trisectors Diagram */
import olympiad;
import olympiad;
size(250);
size(220);


// Equilateral triangle vertices
pair A = (0,0);
pair A = (0,0);
pair B = (10,0);
pair B = (10,0);
pair C = (5,8.660254037844386); // height = 5*sqrt(3)
pair C = (5,8.660254037844386); // height = 5*sqrt(3)


// Draw triangle
// Triangle
draw(A--B--C--cycle, linewidth(1));
draw(A--B--C--cycle, linewidth(1));


// Draw trisectors (hardcoded 20° from sides, ending inside triangle)
// Mark vertices
real trisectorLength = 8.7939; // length of rays inside triangle
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(A--(A + trisectorLength*dir(20)),blue);
draw(C--(C + L*dir(260)), blue+linewidth(0.8));
draw(A--(A + trisectorLength*dir(40)),blue);
draw(C--(C + L*dir(280)), blue+linewidth(0.8));


draw(B--(B + trisectorLength*dir(140)),blue);
// Intersection points (precomputed numerically)
draw(B--(B + trisectorLength*dir(160)),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);


draw(C--(C + trisectorLength*dir(260)),blue);
// Hexagon interior
draw(C--(C + trisectorLength*dir(280)),blue);
filldraw(P1--P2--P3--P4--P5--P6--cycle, lightred+opacity(0.4), red+linewidth(0.8));


pair P1=(4.07604, 3.4202);
// Mark intersection region
pair P2=(5, 4.1955);
label("$P_1$", P1, NW);
pair P3=(5.92396, 3.4202);
label("$P_2$", P2, N);
pair P4=(6.13341, 2.23238);
label("$P_3$", P3, NE);
pair P5=(5, 1.81985);
label("$P_4$", P4, SE);
pair P6=(3.86659, 2.23238);
label("$P_5$", P5, S);
label("$P_6$", P6, SW);


filldraw(P1--P2--P3--P4--P5--P6--cycle, lightred);
label("Hexagon formed by the middle 20° trisectors", (5, -1.2));
</asy>
</asy>
~Avs2010
~Avs2010
~i_am_not_suk_at_math


==Solution 1==
==Solution 1==
Line 45: Line 68:
Using a side of the triangle as a base, we have an isosceles triangle with two <imath>20^\circ</imath> angles. Using this we can show that the third angle is <imath>140^\circ</imath>.  
Using a side of the triangle as a base, we have an isosceles triangle with two <imath>20^\circ</imath> angles. Using this we can show that the third angle is <imath>140^\circ</imath>.  


Following that, we use the vertex angles to show that one angle of the hexagon is <imath>140^\circ</imath>. And with rotational symmetry, three.
Following that, we use the principle of vertical angles to show that one angle of the hexagon is <imath>140^\circ</imath>. And with rotational symmetry, three.
The average of all 6 angles has to be <imath>120^\circ</imath>, so the answer is <imath>\boxed{\textbf{(C) }100}</imath>
The average of all 6 angles has to be <imath>120^\circ</imath>, so the answer is <imath>\boxed{\textbf{(C) }100}</imath>
- SpectralScholar
- SpectralScholar
==Solution 1.5==
Note that the hexagon’s interior angle sum is <imath>720</imath>, and due to 3-way symmetry there are 3 wider(bigger) angles and 3 smaller ones. We can split these angles into 3 groups, each with one big angle and one small angle that sum up to <imath>720/3 = 240</imath> degrees. Seeing this, we can eliminate answer choice <imath>(E)120</imath> as it is asking for the smallest angle. Plugging in the other answer choices yields that only option C with a big angle of <imath>140</imath> and a small angle of <imath>100</imath> keeps the adjacent triangle’s interior angles sum equal to <imath>180</imath>. So the answer is <imath>(C)100</imath>
(Draw a diagram and draw supplementary angles to figure out the sum of the adjacent triangle’s interior angles.)
-Amon26


==Solution 2==
==Solution 2==
Line 58: Line 87:
Note: this is a super informal way to do this, use only if you can't draw a picture or have no idea.
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
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 <imath>AEB</imath> is an isosceles triangle
because the bottom angles A and B are congruent and both <imath>20^\circ</imath>.
Therefore, angle E is <imath>140^\circ</imath>, and the vertical angle in the hexagon
is also <imath>140^\circ</imath>.
Now find G. Triangle <imath>CJD</imath> is isosceles with angles <imath>C</imath> and <imath>D</imath> being <imath>80^\circ</imath>
because angle J in that triangle is <imath>20^\circ</imath>.
Now angles <imath>C</imath>, <imath>D</imath>, and <imath>E</imath> are known and sum to
<imath>80 + 80 + 140 = 300</imath>.
The pentagon <imath>DCFE</imath> and its other vertex (not named in my image)
sum to <imath>540^\circ</imath>.
So subtracting angles <imath>C</imath>, <imath>D</imath>, <imath>E</imath>, and knowing that <imath>F</imath> (let it be <imath>x</imath>)
is congruent (due to symmetry) to the other vertex angle (not named in my image), 
we have:
<cmath>x + x = 240 \implies x = 120^\circ.</cmath>
Thus, angle <imath>G</imath> is <imath>100^\circ</imath> because of triangle <imath>FGE</imath>.
Now find H. In isosceles triangle <imath>AHB</imath>,
angles <imath>A</imath> and <imath>B</imath> are <imath>40^\circ</imath>, so angle <imath>H</imath> is <imath>100^\circ</imath>.
Now find I. The red hexagon’s interior angles sum to <imath>720^\circ</imath>,
and angle <imath>I</imath> is congruent to the angle across from it by symmetry. 
Let <imath>I</imath> and its symmetric angle be <imath>x</imath>. 
Then:
<cmath>2x + 140(E) + 2(100)(G\ \&\ \text{its symmetry}) + 100(H) = 720</cmath>
<cmath>\implies x = 140^\circ.</cmath>
The smallest angle is <imath>100^\circ</imath>.
~PUER_137
==Solution 5 [Most outer triangle]==
Using the outside triangle made by a trisection, we know that two of the angles are <imath>20^\circ</imath> and <imath>60^\circ,</imath> it follows that the third angle in the triangle, the foot of a trisection is <imath>100^\circ.</imath>
We then take a different triangle, that utilizes two of the same lines as the first triangle we examined and also has the <imath>100^\circ</imath> angle. This time, we can use the <imath>40^\circ</imath> angle made by two of the trisections, and we get a triangle with angles <imath>40^\circ, 40^\circ, 100^\circ.</imath>
We can look at a dart-like figure (inverted kite) and we get by symmetry, the angle opposite of the initial <imath>40^\circ</imath> angle is also <imath>40^\circ,</imath> there is also the middle angle formed by the trisection, <imath>20^\circ.</imath> 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 <imath>100^\circ</imath> and by symmetry, that is the smallest angle, so the answer is <imath>\boxed{\textbf{(C) }100}</imath>
~happyfish0922
==Video Solution==
https://youtu.be/l1RY_C20Q2M
==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) ==
== Video Solution (Done in 1 Min) ==

Latest revision as of 19:37, 10 November 2025

Problem

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

~i_am_not_suk_at_math

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 principle of vertical 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 1.5

Note that the hexagon’s interior angle sum is $720$, and due to 3-way symmetry there are 3 wider(bigger) angles and 3 smaller ones. We can split these angles into 3 groups, each with one big angle and one small angle that sum up to $720/3 = 240$ degrees. Seeing this, we can eliminate answer choice $(E)120$ as it is asking for the smallest angle. Plugging in the other answer choices yields that only option C with a big angle of $140$ and a small angle of $100$ keeps the adjacent triangle’s interior angles sum equal to $180$. So the answer is $(C)100$ (Draw a diagram and draw supplementary angles to figure out the sum of the adjacent triangle’s interior angles.)

-Amon26

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

Video Solution

https://youtu.be/l1RY_C20Q2M

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.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2025_AMC_10A_Problems/Problem_6&oldid=268421"