Math Contest! – How long is the RED line?

In this math video I (Susanne) explain how to find the perimeter of the figure. This was a problem from a math contest. Mathematics explained.

00:00 Intro – Math contest
2:15 Solution
6:02 See you later!

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[chrisbaeyens]

My approach is usually letting you solve it and giving you a thumbs up as a reward 😉

[gianluca.g]

I made a massive system of equations, where each variable is a line and each equation is a perimeter. By massaging the equations over and over I got to the solution

[KA-jqet]

Even simpler - Add the 3, 7, 10 and take away the internal boundary, 4 = 16. Add all of the outer regions = 42 and subtract the 16 = 26.

[simon-meadows]

I did it in 3 steps:
1) Paused the video so I could read the problem
2) Stared blankly at it for several minutes, trying to work out how to start
3) Eventually unpaused the video and watched your solution

[blushingmushroom]

I did it the same way but in reverse. I noted the center perimeter of 4 was not needed for the middle perimeter of (3+7+10), making that perimeter (3+7+10)-4=16, then followed that with the outer perimeter not needing that 16 so it became (12+11+9+4+6)-16 = 26.

[DenisMathSolver]

My sincere congratulations to Susanne as a math teacher, for her art of teaching and the ingenuity of her exercises. For me, as an engineer and mathematician, she is among the best on YouTube worldwide.

[speelky]

My solution was: if we sum up all of perimeters, that’ll count all black lines twice. So All black lines can be summed using three regions: black=3+7+10=20. So red=66-2*20=26 🤗

[bpark]

I worked from the center out. Notice that the 4 is totally surrounded by the 3, 10, 7, & the outer circumference includes none of these. So the circumference of the 4/3/10/7 area is 3+10+7 - 4 = 16. The circumference of the whole figure is the sum of the outer circumferences of the outer regions (11+9+4+6+12 = 42) minus the inner circumference (16): 42 - 16 = 26.

[amarug]

my immediate instinct was let linear algebra do the thing, which obviously would have worked but then i remembered it was a 9th grade exercise and that there must be something built in to make it extra easy in terms of calculation. your solution was really elegant

[robertpearce]

Neat solution. I started to do it by working from the inside out
a+b+c=4, etc.
That soon got tedious, and I gave up. Fortunately, I am way past the age of taking any maths exams.

[jeff]

I've been meaning to start pausing your videos and trying to solve the problems on my own, but this one is just so unique I didn't know where to begin.

[שלמהשור]

Amazing, thank you for elegant solution.

[erikheymann]

I added the interior blocks (10, 3, 7, and 4) and subtracted 4, leaving 16, giving you the perimeter of the inner shape. The sum of the perimeters of the outer blocks (12, 11, 9, 6 and 4) is 42. Subtract the inner perimeter and you get 26.

[BrianHaddad]

I think it is impossible to watch one of your videos without smiling. 😊 Thank you!

[sinistre]

The problems are very interesting and your explanations are very clear. This channel is a great discovery for me! I enjoy it!

[TruthTF]

You made it look much easier than it looked. Well done! Btw, loved the hair. Very cute

[davehumbarger]

I love your content and the way you solve the problems. The math is not complex, but the way it is used to solve the problem is the tricky part. You have a great formula to keep us interested. Math has always been a strength of mine, not so with my wife (she'll tell you the same), but we both really enjoy your videos. Thank you!

[LawpickingLocksmith]

OMG! You have melted my brain. Counted all around and then halved it but gon bust doing so!

[samuelweir]

After thinking about how to utilized the information about the inner boundaries, I basically solved the problem in the same way that you did except I started from the inside rather than from the outside of the object. The innermost element is that triangular element label "4" in the middle of the object. From that I determined that the object which was the next layer up had a perimeter of 16, and then the layer up from that had a perimeter of 26.

[moulinaie]

I started like you did at 1:35, giving values of 1 to the sides of the region "3", and then I calculated the other sides, when there was several possibilities, I put "1" on one side and calculate the other. It works very well, and I got the 26 !