 Thanks to Ashish for the suggestion!

Triangle ABC has AB = BC = 10 and AC = 10√2. The triangle is folded so that point A falls on the midpoint M of BC. The crease endpoints P, Q are along sides AB and AC, respectively. What is the length of the crease PQ? As usual, watch the video for a solution.

Folded Triangle Puzzle

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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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(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 may be many ways to solve this problem, but I want to illustrate a solution with coordinate geometry. Why? Coordinate geometry is vital for computer graphics and animation, transforms geometry questions into algebraic equations, is useful is higher level mathematics, and is a powerful problem-solving technique. We can simply make a few equations and solve the problem like magic!

First notice:

102 + 102 = (10√2)2
AB2 + BC2 = AC2

Thus ABC is a right triangle with hypotenuse AC. Set a coordinate system with B = (0, 0), and let A be 10 units up at (0, 10) and C be 10 units to the right at (10, 0). Then the midpoint of BC is M = (0, 5). Label points P along AB and Q along AC. We don’t know the coordinates of P, but we can say it has an x coordinate equal to 0. So let its coordinates be (0, p). The segment AP is folded into PM, so clearly the two segments have the same length. This means:

|AP| = |PM|
10 – p = √((5 – 0)2 + (0 – p)2)
(10 – p)2 = 52 + p2
100 – 20p + p2 = 25 + p2
20p = 75
p = 15/4

So we have P = (0, 15/4). Let’s work out the coordinates of Q.

Method 1: geometric principles

We could use some geometric principles to simplify the calculation. Just as |AP| = |PM|, we also know |AQ| = |QM|. This means the points P and Q are equidistant from A and M, so PQ is the perpendicular bisector of AM. We could calculate AM has slope -10/5 = -2, and therefore PQ must have the negative reciprocal slope of 1/2. Since P = (0, 15/4), we can use the slope-intercept form to get:

PQ line
y = mx + b
y = (1/2)x + 15/4

We can readily solve for the line AQC:

AQC line
y = mx + b
Using points A(0, 10), C(10, 0)
b = 10
m = (0 – 10)/(10 – 0) = -1
y = –x + 10

Q is the intersection of the two lines, so we have to solve:

y = (1/2)x + 15/4
y = –x + 10

Subtracting gives:

0 = (3/2)x – 25/4
x = 25/6

Substituting back

y = -25/6 + 10 = 35/6

We now just find the distance between coordinates P(0, 15/4) and Q(25/6, 35/6). This gives us:

|PQ|
= √((25/6 – 0)2 + (35/6 – 15/4)2)
= 25(√5)/12
&approx; 4.658

Method 2: distance formula again

What if we didn’t remember that geometric insight about equidistant points? In a normal geometry problem you might be stuck. But with coordinate geometry, you may just be able to hack your way to the answer!

Just as in method 1, we find the equation for the line ACQ:

AQC line
y = mx + b
Using points A(0, 10), C(10, 0)
b = 10
m = (0 – 10)/(10 – 0) = -1
y = –x + 10

Suppose Q has an x coordinate equal to q. Then its y coordinate will be –q + 10. We now set the distances AQ and QM equal to each other. It will be convenient to square the distances to avoid the square roots.

A = (0, 10)
Q = (q, –q + 10)
M = (5, 0)

|AQ|2 = |QM|2
(q – 0)2 + (-q + 10 – 10)2 = (5 – q)2 + (0 – (10 –q))2
2q2 = 2q2 – 30q + 125
30q = 125
q = 25/6

Now we have –q + 10 = 35/6, so we again get Q = (25/6, 35/6).

As before, we now just find the distance between coordinates P(0, 15/4) and Q(25/6, 35/6). This gives us:

|PQ|
= √((25/6 – 0)2 + (35/6 – 15/4)2)
= 25(√5)/12
&approx; 4.658

And that’s it! We could solve the problem without remembering every little theorem and proposition from geometry class. Coordinate geometry is so powerful that it almost feels like cheating!

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