Aug 23, 2022

# Magical Triangle - Think Outside The Box!

mind your choices i am presh tallwalker a b c d is a square with sides equal to 1. pick point e next to dc and pick point f next to bc so that e a f is a 45 degree angle and a fb is a 70 degree angle is the first part of the question is the solution to the measure of the angle aef the second part of the question is the solution to the perimeter of the

### triangle

ecf this is solution to ec plus cf plus fe a version of this problem was given to ninth grade students in india who had only learned geometry , so the challenge is to solve it without trigonometry.
Pause if you'd like to try this issue and if you're willing to keep watching to learn how to solve this we're going to solve it by

### think

ing outside the box I give On and Gotham the credit for the solution she emailed me. Now let's start. Let's first consider the

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abf, since angle b is a 90 degree angle angle dae must equal 25 degrees, since angle d is a 90 degree angle angle aed must equal 65 degrees. Now let's go back to

### triangle

abf. The trick to this problem is to

### think

outside the box, we rotate this

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90 degrees clockwise.
Through vertex a we have created a congruent

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and we call this vertex f prime because it is the image of vertex f. From here we have f prime a d is equal to 20 degrees by design. Then we will solve that angle f prime a e is equal to 20 degrees plus 25 degrees, which is a total of 45 degrees. Now we will consider that a f prime is equal to af by construction. So we have two

### triangle

s that are congruent through the side angle side here one side that is equal then we have two angles that are equal to 45 degrees and then we have two sides that are equal to each other so these two

### triangle

s are congruent , there are actually mirror images of each other, which means that the angle aef exactly equals an angle aef prime and so it equals 65 degrees and we're done with the first part of the question.
Now let's solve the second part of the question, what is the perimeter of the

### triangle

ecf. Let's consider a numerical approach, imagine we have a ruler and that we can measure the length of ec. Let's say we estimate it to be 0.6389. Let's do the same for the other sides of cf and fe. In the end it turns out that the sum of these three lengths is equal to two. What would happen if we change the points e and f, something remarkable happens, no matter where we pick the points of e and f, the perimeter of this

### triangle

will always be equal to 2, which is just good enough for a numerical approach, but why does it happen mathematically, let's prove that So let's go back to our original diagram, construct

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af prime number d.
Recall that

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f prime e is congruent to

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afe. Let's label some lengths. First assume that b f has a length equal to x. This means that d f prime now also has length equal to x fc must have length one minus x since the entire side bc has length one. Next we assume that d e has a length of y, then we must have f prime e that has a length of x plus y will have the same length as ef so ef has a length equal to x plus y finally ec has a length equal to 1 minus y because the entire side dc has a length equal to one now let's look at the perimeter of the

### triangle

ecf it will be equal to one minus y plus one minus x plus x plus y we can cancel the y's and the x's and we have one plus one and that equals exactly two what we saw in numerical approach amazing thanks for considering your choices one of the best channels on youtube as always thanks for watching and thanks for your support you
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