mind your decisions i'm presh tallwalker
a b c d is a square with a side length
equal to 1.
pick point e alongside dc and pick point
f alongside bc such that e a f is a 45
degree angle and a fb is a 70 degree
angle
[Music]
the first part of the question is to
solve for the measure of angle aef
the second part of the question is to
solve for the perimeter of triangle ecf
that is solve for ec plus cf plus fe
a version of this problem was given to
ninth grade students in india who had
only learned geometry
therefore the challenge is to solve
without using trigonometry
pause if you'd like to give this problem
a try and when you're ready keep
watching to learn how to solve this
problem
[Music]
we will solve the problem by thinking
outside the box
i give credit to the solution to on and
gotham who emailed it to me
now let's get started
first consider triangle abf
since angle b is a 90 degree angle
angle fab must be equal to 20 degrees
then consider triangle ade
since angle a is a 90 degree angle angle
dae must be equal to 25 degrees
then since angle d is a 90 degree angle
angle aed must be equal to 65 degrees
now let's go back to triangle abf
the trick to this problem will be to
think outside the box
we will rotate this triangle
90 degrees clockwise about vertex a
we've created a congruent triangle and
we'll label this vertex as f prime
because it is the image of the vertex f
from here we have f prime a d is equal
to 20 degrees by construction
we then will solve that angle f prime a
e will be equal to 20 degrees plus 25
degrees which is equal to a total of 45
degrees
now we are going to consider that a f
prime is equal to af by construction
we thus have two triangles that are
congruent by side angle side
here we have one side that's equal to
itself
then we have two angles that are equal
to 45 degrees and then we have two sides
that are equal to each other
therefore these two triangles are
congruent
in fact there are mirror images of each
other
this means angle aef is exactly equal to
angle aef prime and so it is equal to 65
degrees and we're done with the first
part of the question
now let's solve for the second part of
the question
what is the perimeter of triangle ecf
let's consider a numerical approach
imagine that we have a ruler and that we
can measure the length of ec
suppose we estimate it to be 0.6389
let's do the same thing for the other
sides of cf and fe
we end up that the sum of these three
lengths will be equal to two
now what would happen if we change the
points e and f
a remarkable thing happens no matter
where we pick the points of e and f the
perimeter of this triangle will always
be equal to 2 exactly
this is good enough for a numerical
approach but why does it happen
mathematically
let's prove this so we go back to our
original diagram
construct triangle af prime d
recall that triangle f prime e is
congruent to triangle afe
let's label some lengths
first suppose that b f has a length
equal to x
this means d f prime also has a length
equal to x
now fc must have a length equal to one
minus x because the entire side bc has a
length equal to one
next suppose that d e has a length equal
to y
we must have that f prime e which has a
length of x plus y will have the same
length as ef
therefore e f has a length equal to x
plus y
finally ec will have a length equal to 1
minus y because the entire side dc has a
length equal to one
now let's consider the perimeter of
triangle ecf
it'll be equal to one minus y plus one
minus x plus x plus y
we can cancel out the y's and cancel out
the x's and we're left with one plus one
and that's exactly equal to two which is
what we saw in the numerical approach
amazing
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