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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.This entry contributed by Alejandro Astudillo.Click the figure below to see the complete problem 674.
Thank you Antonio to put it in your site, this a really hard problem!
AH:CH = tan 5x : tan 3x and alsoAH:CH = tan 2x : tan x tan 5x tan x = tan 3x tan 2x sin 5x cos 3x cos 2x sin x = cos 5x sin 3x sin 2x cos x (sin 8x + sin 2x)(sin 3x – sin x) = (sin 8x - sin 2x)(sin 3x + sin x)sin 8x : sin 2x = sin 3x : sin x (by componendo & dividendo)sin 8x sin x = sin 3x sin 2xsin 8x = 2 sin 3x cos x = sin 4x + sin 2xsin 8x – sin 4x = sin 2x2 cos 6x sin 2x = sin 2xcos 6x = 1/2 6x = 60 degx = 10 deg
Of course, the trigonometry version is pretty easy, how about a synthetic solution?
If trigonometry be allowed then even Ceva's Theorem in trigonometric form directly leads to sin(10x)= 4cos(x)sin(3x)cos(6x) and thence to sin(8x)= sin(2x)+ sin(4x) or cos(6x)=1/2 whence x=10. But the real challenge, I guess, is to find an elegant plane geometry solution. In this regard, I only have a suggestion to make: Extend CD to meet AB in E. Now /_ADE=3x while /_EAD=90-6x. If we can now prove that DE=AE then 90-6x=3x or x=10. However, so far I've not been successful in proving that Tr.ADE is isosceles.Ajit
The altitude of triangle ABC is BH, not BD.
Extend HC rightwards and construct AB=BI, I is on extension of HC. Connect DI; extend AD rightwards to intersect with BC at J; make DK intersect at AC so Angle CDK=Angle DAC;and connect JI.
P.S. Connect JK.
Since AB=BI in Iso Tri ABI, AD=DI (Angle DAI=DIA=x), Angle DCA-Angle DIC=Angle CDI=x, DC=CI. Angle JDC=Angle DBC=x, Angle JCD=DCB, Tri JDC~Tri DBC, DC^2=JC*CB; CI=DC, CI^2=JC*CB, Tri CJI~Tri CIB, Angle CIJ=Angle CBI=2x.Angle JID=Angle JIA-Angle DIA=x;DI bisects Angle JIA. Angle CDK=Angle CAD=x, Angle DCK=Angle ACD, Tri CDK~Angle CAD, DC^2=KC*AC. (ID bisects Angle JDK.)JC*CB=KC*AC, cyclic quadrilateral ABJK, Angle JKI=8x. 8x+x=90, x=10.