Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Sunday, January 18, 2015
Geometry Problem 1074: Quadrilateral, Right Triangle, Angles, 15, 22,5, 30, 90 Degrees
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Let O is a circle through A, B and D where cuts AC and DC at E and F respectively.
ReplyDeleteLet AD and BC intersect at G
Note that A, B, E, F and D concyclic
Because ∠GBA an exterior angle of triangle ABC then m ∠GBA = 52.5 degree
so that m∠DBC = 105 degree. This implies ∠DBC = 30 degree
Because BADF cyclic then m∠BAF = 30 degree so that m∠EAF = 7.5 degree = m∠EDF (A, B, E, F and D concyclic)
Because ∠AED an exterior angle of triangle ECD then m ∠AED = (7.5 + 15) degree = 22.5 degree = x
(From wisna email: putu_wisna_ariawan@yahoo.com)
To Anonymous
ReplyDeleteRefer to line 5 "so that m∠DBC = 105 degree"
Please provide more explanation how do you get 105 degree.
http://s17.postimg.org/mmbwojrcv/pro_1074.png
Peter Tran
To Peter Tran
ReplyDeleteI am sory because there was an error in my solution in line 5.
Thanks for your corection. But I am sure that x = 22.5 degree.
I still working to find that solution.
With out loss of generality assume AB = 1.
ReplyDelete(all angles are measured in degrees)
First note CAD = 67.5,
ADC = (180 - 82.5),
ABC = (180 - 57.5)
By Sine Rule applied to Δ ABC:
AC / sin 57.5 = AB / sin 30 = 2.
So AC = 2 sin 57.5 = 2 cos 37.5 ..... (i).
By Sine Rule applied to ΔADC:
AD / sin 15 = AC / sin 82.5 = AC / cos 7.5
So AD = AC. [sin 15 / cos 7.5]
= 2 [cos 37.5].[2 sin 7.5]
= 2[2 cos 37.5sin 7.5]
= 2[sin 45 – sin 30]
= 2[1/(sqrt 2) - 1/2]
= (sqrt 2) - 1 = tan 22.5
So tan x = AD / AB = (tan 22.5) / 1 = tan 22.5
Hence x = 22.5 degrees
Let O be the center of a circle through B, D and C where cuts AC at P so we have triangle BPO equilateral (I) ( Because m∠BOP = 2m∠BCP=30º.Now have m∠BOD = 2.m∠BCD = 90º so #BODA is cyclic because m∠BAD + m∠BOD = 90º+90º= 180º then m∠BAO = m∠BDO = 45º but m∠BAO = m∠BAC+m∠CAO then m∠CAO = 22.5º (II). Circunference ABOD cuts AC at Q but AC bissects ∠BAO then we have BQ = QO then using (I), triangles BPQ and QPO are congruents by equal sides then m∠BPQ = 30º = m∠BAP + m∠ABP ( ∠BPQ external angle) .: m∠ABP = 30º - 22.5º = 7.5º (III) but m∠PBD = m∠PCD = 15º (inscritive angles), finaly using (III) we have m∠ABD = m∠ABP + m∠PBD = 7.5º + 15º .: m∠ABD = 22.5 º
ReplyDeleteLet O be the circumcentre of ABC and let CO and AD meet at E.
ReplyDelete< AOB = 60 so Tr. OAB is equilateral. Also < BOC = 45 so < AOE = 75. Hence <OAC = OCA = 37.5. Hence < DCE = 22.5 and < OAE = 30 so it follows that AO = AE
Therefore Tr. BAE is right isoceles and so < AEB = 45 = < DCB implying that BCED is cyclic.
So < DBE = < DCE = 22.5 and so x = 22.5
Sumith Peiris
Moratuwa
Sri Lanka
There are more general forms of this question.
ReplyDeleteThe first one is, BAD = 120-4t=90, BAC=30-t=22.5, BCA=30, ACD=2t=15 where t=7.5 and the result is 3t=22.5.
The second one is, BAD = 60+4t=90, BAC=3t=22.5, BCA=60-4t, ACD=30-2t=15 where t=7.5 and the result is 30-t=22.5
I have not investigated yet that the above solutions are also the solutions of the general form. One can also try to give the solution for the general cases.