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In number theory, many problems are easy to state but difficult to solve within number theory.

Are there Elementary (solved and unsolved) problems from Euclid's geometry, which are difficult to solve only by Euclid's axioms or the newest Hilbert's axioms on geometry? Had Euclid or Archimedes posed such propositions which they couldn't prove?

(The most well known such questions are from Ruler and compass constructions, and doubling cube. I would like to see the problems, whose statements are like the propositions of Euclid, but whose proof requires non-geometry tools.)

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I guess proving that Morley's triangle is equilateral would qualify as "difficult to solve only by Euclid's axioms".

https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem

In plane geometry, Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, called the first Morley triangle or simply the Morley triangle.

Some more pointers at http://mathworld.wolfram.com/FirstMorleyTriangle.html.

[ EDIT ] Another example is Monsky's theorem https://en.wikipedia.org/wiki/Monsky%27s_theorem.

In geometry, Monsky's theorem states that it is not possible to dissect a square into an odd number of triangles of equal area. In other words, a square does not have an odd equidissection.

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enter image description here

One well-known problem that is challenging is a "sangaku":

In a circle $O$ with diameter $AB$, draw another circle $Q$ internally tangent to $O$ at $A$, which intersects $AB$ at point $P$. on $PB$ construct an isosceles triangle $\triangle BCP$ so that $BC = CP$. Inscribe a circle with center $I$ such that it is internally tangent to circle $O$, externally to circle $Q$, and to $CP$. Prove that $IP \perp AB$.

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  • $\begingroup$ Where has it been used that $\triangle BCP$ is isosceles? $\endgroup$ – Christian Blatter Nov 21 '15 at 14:52
  • $\begingroup$ @ChristianBlatter The triangle must be isosceles because otherwise you could choose any point $C$ on the semicircular arc $AB$. $\endgroup$ – heropup Nov 21 '15 at 17:12
  • $\begingroup$ From your animation it now becomes apparent that $C\in O$, which was not stated in the question. $\endgroup$ – Christian Blatter Nov 21 '15 at 18:43

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