Looks Easy But Is Insanely Hard

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Many people asked me to solve this one! What is the value of x?

Many more solutions

Solution by Ahmet ÇETİN

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Pretty simple to solve. After you perform the blood sacrament at the temple of Archimedes to divine where to draw extra lines. Very simple

KUWAITGRIPSVEVO
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I really need to start thinking outside of the triangle

theimmux
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A straightforward trygonomeyric solution, most likely not what required/expected:
BP extended until it crosses AC in D will hit AC at 90 degrees, so angle APC is
a sum of angles DPC and DPA which is
+ 30 degrees

vladimirrainish
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We can use similar triangles here:
at 0:37,
Continue the line BP to D in order to intersect AC a D.This is the height of triangle ABC since AB=BC.
Angle ABC=Angle ACB=(180-20-60)/2 =50 deg.
Consider triangle PAD,
This is 30, 60, 90 with AD=1, AP=2 and PD= sqrt3 say,
Create a mirror image triangle ACF similar to triangle BAD to intersect AC at E.
We now have an isosceles triangle ABE with two lines of length BC a base of 1+1=2 and angles 80, 80 and 20 degrees.
Draw a similar triangle CFA .Let G be the centre point of FA.GA will be the height of triangles AFC and ABC.
Let BE and FC intersect at H and assume for the moment that triangle BAE= triangle CFA.
BH=HC since triangle CFA= triangle ABE
Triangle BHC is isosceles.
angle HBC=angle HCB=(50-10-10))=30 deg
angle BPC=180-40-30=110 deg
angle PCA=50-30=20 deg which is the apex of triangle ACF
Now 360=x+angle BPA+BPC=x+150+110.
x=360-150-110
x=100 degrees.
I proved that triangles CBE and CFA were equal and triangle PFA was similar to CBE and CFA by trigonometry methods which reinforced my original conjectures.

shadrana
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Heyt be bu kanalı benden başka türklerde takip ediyormuş sevindim. Ahmet seviliyorsum

BahadR
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Presh: provided a complicated solution with drawing many additional lines
Me: I think there are must easier way

LogicalMath
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Found a right triangle (not the same one as presh) and used trigonometry to solve for x directly. The final calculation looked quite messy, but it's still well-defined

tomtheultimatepro
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Just think: Some people are watching this video and some people are watching Space Jam on FreeForm right now.

christopherrice
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when i see the problem
first: let's sovle it without pencil
then: kinda complicated, but almost done
a few moments later: i did it on the 15th paper 😎

nvztsnl
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Let F be the point on AE such that angle PBF=10°. Then AP and BP are both bisectors in triangle ABF, so FP bisects angle AFB and angle PFB= angle AFP= 60°. Furthermore, by simple angle chasing we have that points C, F, P are collinear and angle BCF= angle BCP= 30°. And the rest is easy.

АндрейМихалёв-уэ
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Just another way to SOLVE,
Draw a circle making A the centre, as the triangle is isosceles it must pass through B, C. Now extend AP to the end of the circle, let the point be D. as DAC=60, AC=AD=DC
Also extend BP to the end of the circle, let this point be F.
Consider traingle BDC and APF, you will easily find they are congruent and PF=BC
Now angle BFC=80÷2 =40(Central angle is 2 times the angle made with circumference) angle CBF is also 50-10=40, so BCF is isosceles with BC=CF, thus PF=CF and angle PFC 40 so angle FPC=180-40/2=70, notice ACB=50 degrees and PBC=40, So BF is perpendicular to AC thus angle APF=90-60=30, required angle = APF+CPF=30+70=100 (ANS)

Mathematician
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Sir, how can we conclude that a given problem required a construction

kirancreations
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I really like this problem. I did find the symmetry axis AD but I couldn't figure out all angles purely from that but I did find calculation path via the lengths - first calculating AP and then CP (as multiples of AC). The last arcsin calculation then gives 2 possible solutions for x (80 and 100 degrees) - but only the 100 degrees (which does not appear on the calculator screen) is the right one. Nice and tricky!

ericschmidt
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This is very simple problem... You solved it in a very complex way...
Not the first time I'm seeing an unnecessary complicated solution on this channel...

chotu
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تمرين جميل . وشرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

اممدنحمظ
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About 2 years ago i saw this question while i was preparing for University exams in Turkey. I tried to solve too much, for example ı tried to cut a triangle an paste it near another one (usually hard problems can be solved in this way) but they never worked. Finally one of my teachers solved this problem
I don't really remember if he used this way to solve but it was similar at least. Its good to see the solving of the problem again. Thanks.

arifemre
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Sir I request you to do some tough secondary maths questions too on logs, exponentials, calculus, functions, limits, continuity, etc

thecoolring
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Clever solution!! I did it with the Trigononometric version of Ceva's theorem, a bit messy but pretty straightforward!

argyrisgiannisismanes
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I tried to solve this problem using trigonometry, and I found x in four different forms, all involving trigonometric and inverse trigonometric functions (e.g. arcsin(1-sin(...)/sin(...)). I was able to find the value of 100 degrees using the calculator, but once I found it I was able to rigorously prove it. Does this count as a valid, authentic solution? I wonder.

adammaths
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I just used the Sine rule on both the equal sides with angles P, B and C, and substituted one into the other and solved for x.

thdoctor