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.

Or keep reading.

.

.

“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.

.

.

.

.

.

.

M

I

N

D

.

Y

O

U

R

.

D

E

C

I

S

I

O

N

S

.

P

U

Z

Z

L

E

.

.

.

.**Answer To Folded Triangle 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 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:

10^{2} + 10^{2} = (10√2)^{2}*AB*^{2} + *BC*^{2} = *AC*^{2}

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} = 5^{2} + *p*^{2}

100 – 20*p* + *p*^{2} = 25 + *p*^{2}

20*p* = 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

≈ 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}

2*q*^{2} = 2*q*^{2} – 30*q* + 125

30*q* = 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

≈ 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!

### MY BOOKS

If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.

Book ratings are from January 2023.

(US and worldwide links)

https://mindyourdecisions.com/blog/my-books

**Mind Your Decisions** is a compilation of 5 books:

(1) The Joy of Game Theory: An Introduction to Strategic Thinking

(2) 40 Paradoxes in Logic, Probability, and Game Theory

(3) The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias

(4) The Best Mental Math Tricks

(5) Multiply Numbers By Drawing Lines

**The Joy of Game Theory** shows how you can use math to out-think your competition. (rated 4.3/5 stars on 290 reviews)

**40 Paradoxes in Logic, Probability, and Game Theory** contains thought-provoking and counter-intuitive results. (rated 4.2/5 stars on 54 reviews)

**The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias** is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 4.1/5 stars on 33 reviews)

**The Best Mental Math Tricks** teaches how you can look like a math genius by solving problems in your head (rated 4.3/5 stars on 116 reviews)

**Multiply Numbers By Drawing Lines** This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 4.4/5 stars on 37 reviews)

**Mind Your Puzzles** is a collection of the three “Math Puzzles” books, volumes 1, 2, and 3. The puzzles topics include the mathematical subjects including geometry, probability, logic, and game theory.

**Math Puzzles Volume 1** features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.4/5 stars on 112 reviews.

**Math Puzzles Volume 2** is a sequel book with more great problems. (rated 4.2/5 stars on 33 reviews)

**Math Puzzles Volume 3** is the third in the series. (rated 4.2/5 stars on 29 reviews)

### KINDLE UNLIMITED

Teachers and students around the world often email me about the books. Since education can have such a huge impact, I try to make the ebooks available as widely as possible at as low a price as possible.

Currently you can read most of my ebooks through Amazon’s “Kindle Unlimited” program. Included in the subscription you will get access to millions of ebooks. You don’t need a Kindle device: you can install the Kindle app on any smartphone/tablet/computer/etc. I have compiled links to programs in some countries below. Please check your local Amazon website for availability and program terms.

US, list of my books (US)

UK, list of my books (UK)

Canada, book results (CA)

Germany, list of my books (DE)

France, list of my books (FR)

India, list of my books (IN)

Australia, book results (AU)

Italy, list of my books (IT)

Spain, list of my books (ES)

Japan, list of my books (JP)

Brazil, book results (BR)

Mexico, book results (MX)

### MERCHANDISE

Grab a mug, tshirt, and more at the official site for merchandise: **Mind Your Decisions at Teespring**.