Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Erina-NJ.
Click the figure below to see the complete problem 920.
Sunday, September 8, 2013
Geometry Problem 920: Triangle, Angles, 10, 20, 30 Degrees, Congruence
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I think that Ceva's trigonometric theorem gives:sin(30)/sin(10)*sin(x)/sin(120-x)=1 which yields, x=20° and /_BCD=100° as may ascertained from the diagram above.
ReplyDeleteIn any case, what we are looking for is a synthetic solution, aren't we?
(Thank you once again Antony Gutierrez)
ReplyDeleteProblem 920. Pl refer to my earlier post:
By Ceva,
sin DAB. sin DBC. sin DCA = sin DAC. sin DBA. sin DCA
sin 10°.sin 30°.sin x = sin 10°.sin 10°.sin(120°-x).(This line is corrected)
So sin 30°.sin x = sin 10°.sin(120° - x).
Since sin 30° = 1/2 and sin(120°-x)= Sin(90°+30°-x)= cos(30°-x)
sin x = 2 sin 10°.cos(30°-x)
(OR)
By sine rule:
sin x / sin 10° = AD/DC = DB/DC = sin(120°-x)/sin 30° = 2 sin(120°-x)
So sin x = 2 sin 10°.sin(120°-x)= 2 sin 10°.cos(30°-x)
By inspection x = 20° satisfies the above equation.
P.S.:Is there a simple way of solving the above trig equation?
Problem 20 From Erina Nj To Pravin
DeleteSinx = 2 sin10 cos (30-x)
sinx = sin (40-x)- sin (20-x)
sin (40-x)- sinx= Sin (20-x)
2sin (20-x) cos 20-sin (20-x) =0
sin(20-x) [2cos20-1] =0
2cos20-1 different from 0 => sin(20-x) = 0
X=20
@Anonymous, how about a synthetic solution to this exvellent problem?
Deletehttp://www.mathematica.gr/forum/viewtopic.php?f=22&t=40350
ReplyDeletecan you solve using elemtery geomitry
ReplyDeleteMANOLOUDIS APOSTOLIS
Delete4th HIGH SCHOOL OF KORYDALLOS -PIRAEUS-GREECE
PROBLEM 920
SOLUTION 1
E is a point of AC. Then <EBC=10 degrees so <DBE=20 degrees .If O the pericenter of triangle ABE then OA=OE=OB=AE because <AOE=2×<ABE=2×30=60 degrees and <OAB=<OBA= 40 degrees. Also DA=DB so OD is perpendicular to AB and OD is vertical to the middle of AB hence <AOD=50 degrees= <OAD thus DO=DA,OE=AE so tri ADE=tri ODE.
So <DEA = 60/2= 30 degrees = <DBC so DECB is inscribable hence <DCA=<DCE=<DBA= 20 degrees
SECOND SOLUTION
Let E, F the symmetrical of point A to DB and BC respectively. Then DA=DE=DB, AB=BE=BF, <CBF=<CBA= 40 degrees, <CFB= <CAB= 20 degrees. Then B is the pericenter of triangle AEF so <AFE =10 degrees so triangle EBF is equilateral. So DE=DB, EF=BF so DF is vertical to the middle of EB. We conclude that <DFA= 20 degrees, <DFC= 10 degrees= <DAC
so ADCF is inscribable hence <DCA=<DFA= 20 degrees
It seems the following proof it's not among the given ones: Take E the reflection of C in AD; easily BE=CE. Construct the equilateral triangle CED' inside the Langley triangle ACE; D'E=BE and <ABD'=10 degs, hence D'=D. Now <ACD=<AED=20.
ReplyDeleteHappy New Year!
Excellent work Apostolis and Stan in finding these pure geometry proofs
ReplyDeleteThank you very much for your nice words Sumith
ReplyDeleteAnother nice proof using isogonal conjugate point and 'triple angle lemma' converse at https://stanfulger.blogspot.com/2026/04/gogeometry-antonio-gutierrez-problem-920.html
ReplyDelete