User:MRFS/Sandbox

Here is an easy geometric proof based on a construction in the Taylor/Marr paper cited below.

In any triangle ABC let X be the Morley vertex adjacent to BC. Construct the points P and Q on AB and AC respectively such that |BP| = |BX| and |CQ| = |CX|, and then Y on the trisector CS such that ∠XPY is 30°. The foot of the perpendicular from X to PY is R. The six marked segments will all have equal length. Note that the three right-angled triangles, ΔRXY and ΔSXY and ΔSQY, (called wedges) have equal hypotenuses and an equal (marked) side therefore they are congruent. Evidently α + β + γ = 60°


 * ∴  ∠QXP = 360° − 2(90° − β) − 2(90° − γ) = 120° − 2α


 * so ∠YXR =∠SXY =∠YQS = (∠QXP  − 60°)/2 = 30° − α.


 * ∴  ∠QYP = 360° − 3(60° + α) = 180° − 3α


 * ∴  APYQ is a cyclic quadrilateral.


 * Now ∠XPQ = ∠PQX = 30° + α   since ΔPQX is isosceles


 * ∴  ∠YAQ = ∠YPQ  = ∠XPQ − 30° = α


 * ∴  Y is the Morley vertex adjacent to AC.

Likewise the point Z where ∠PBZ = β and ∠ZQX = 30° is the Morley vertex adjacent to AB and is obtained by a similar construction (outlined). This generates three more wedges which are clearly congruent to the first three. In particular |XY| = |XZ| and ∠YXZ = 60° therefore ΔXYZ is equilateral.

There's very little in it but it may be marginally neater to replace the third indented line above to say that QYP=360-3(60+α)=180-3α so APYQ is a cyclic quad.
 * ∴  ∠YPA = 60° + β   and   ∠AQY = (90° + γ) − (30° − α)   are supplementary

This effectively eliminates β and γ from the later part of the proof and potentially makes the symmetry argument a bit clearer.

For better motivation let O be the circle center X that touches BC. ... and then Y on the trisector CS such that PY is tangent to O at R?

Maybe have second thoughts about posting this in the light of the new proof at cut-the-knot.

Here's my latest version, basically the TeX from Direct6 :-

Draw the six trisectors t1, ... ,t6 as shown. X is the Morley vertex where t4 cuts t6. Let P and Q be the points on AB and AC such that BP = BX and CQ = CX, and let t1 and t2 cut the circle through A, P, Q at Z and Y. Then PZ = ZY = YQ as the chords PZ, ZY, YQ subtend equal angles at A. Finally choose R on the opposite side of YZ to X so that ΔRYZ is equilateral.

Now ΔPXR and ΔQXR are congruent since PX = 2HX = QX, RX is common, and PR = QR by symmetry. So ∠QXR = &frac12;∠QXP = 60° - α as the reflex angle ∠PXQ = 2(90° - β) + 2(90° - γ)= 240° + 2α. The circle centre Y and radius YQ also goes through R and Z, and ∠YZQ = ∠YAQ = α so ∠QZR = 60° - α = ∠QXR. So X too lies on this circle and thus ΔQXY is isosceles. Hence t5 actually goes through Y. By the same token t3 passes through Z, which means Y and Z are the Morley vertices adjacent to AC and AB respectively. ΔXYZ is equilateral as PZ = ZX = XY = YQ = YZ.