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Angle In Arch Puzzle – Mind Your Decisions
Here’s a fun problem that is making the rounds.
These puzzles are fun, but in order to be solvable you have to make assumptions about angles and lengths from the diagram. This is problematic because in real life many diagrams are not drawn to scale, so you do not want to be in the habit of making such assumptions.
It is an interesting mental challenge to convert a puzzle into a more precise mathematical statement. I did some research and found this work was already done, as the problem appeared in the 2009 AMC 10B (one of the qualifying tests for the U.S. mathematical Olympiad team).
The AMC problem is stated like this:
A keystone arch, shown below, is composed of 9 congruent isosceles trapezoids fitted together along the non-parallel sides. The bottom sides of the two end trapezoids are horizontal. Let x be the angle measure in degrees of the larger interior angle of the trapezoid. What is x ?
The mathematical statement is more precise, however, it is also more verbose. So it is a challenge of puzzle makers to present problems precisely, but without too much jargon which would alienate the casual puzzle solver.
I thought it was interesting to juxtapose the two presentations of this problem. With that said, let’s work out the answer!
As usual, watch the video for a solution.
Or keep reading.
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Answer To Angle In Arch Puzzle
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
There are many ways to solve this problem. I will present 4 methods that I found interesting.
Method 1: triangles
Extend the non-parallel sides of each trapezoid.
Most people just assume all the line segments meet at a common point, but one should justify this is the case.
Consider two adjacent trapezoids, say the leftmost and the next one. The trapezoids are congruent, so the angles and lengths of the triangle formed by the extended line segments are congruent, so the adjacent diagram is a mirror image about the common line segment. Thus the line segments meet at the same point. Continue this reasoning to each adjacent trapezoid, so all trapezoids meet at the same common point. By symmetry this point will be the center of the horizontal line segment.
We have now formed 9 congruent angles y that form a straight line, so 9y = 180° and y = 20°.
We have also formed 9 congruent isosceles triangles with a vertex angle of y and non-vertex angles of z. So we have 2z + y = 180° and z = 80°.
Finally x + z = 180°, so x = 100°.
Method 2: decagon
Form a decagon with the horizontal line and the 9 periphery sides of the trapezoids.
The sum of the interior angles can be computed by the well known formula with n = 10 to get:
180°(n – 2)
= 180°(10 – 2)
= 1440°
The larger interior angle in each trapezoid is x. Let y be the smaller interior angle in each trapezoid. In a quadrilateral the sum of interior angles is 306°, so 2x + 2y = 360° and y = 180° – x.
The 10 sided polygon has 2 interior angles of y, and the remaining 8 are formed by 2y, so we have a total sum of 2y + 8(2y) = 18y. (We could equivalently see there are 9 trapezoids and each contributes 2y to the interior angle sum, to make 9(2y) = 18y.) Thus we have:
18y = 1440°
y = 80°
Finally we have:
y = 180° – x
100 = 180° – x
x = 100°
Method 3: octadecagon
We can similarly reflect the arch to form a regular 18-sided polygon.
Using the interior angle sum formula we get:
180°(n – 2)
= 180°(18 – 2)
= 2880°
We can also sum the interior angles of the 18-sided polygon in terms of the angles y of the trapezoid:
18(2y = 2880°
y = 80°
Finally we have:
y = 180° – x
100 = 180° – x
x = 100°
Method 4: turning angle
We can also employ exterior angles to solve this puzzle.
Consider the angle z a trapezoid has to be turned to be in the orientation of the adjacent trapezoid. A pen that is turned 9 such times will be rotated upside down 180°.
Thus we have:
9z = 180°
z = 80°
We form a straight line between two angles of z and one of y, so we have:
2z + y = 180°
2(20°) + y = 180°
y = 80°
Finally as before we have:
y = 180° – x
100 = 180° – x
x = 100°
What a fun puzzle!
Special thanks this month to:
Daniel Lewis
Lee Redden
Kyle
Thanks to all supporters on Patreon and YouTube!
References
Puzzle on X
https://x.com/sonukg4india/status/2038832197256585556
2009 AMC 10B Problems/Problem 24 via AoPS
https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_24
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