## Thursday, August 10, 2017

### Geometry Problem 1341: Isosceles Triangle, 80-20-80 Degrees, Circumcenter, Angle Bisector

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. Problem 1341
Suppose the AD crosses the circle at point E, then arcBE=arcEC and <ECB=40, <ECO=50=<OEC (<COE=80).I form the equilateral triangle OCF (F is arcEC ).
Is < FOE=20, if G is circumcenter of triangle FOE then GO=GE=GF and <GOE=10=<GEO.
Is triangle OGC=triangle GFC so <GCO=<GCF=60/2=30.But <FCE=<FOE/2=20/2=10
and <ECG=20.Now triangle DCE=triangle GEC so DC=GE.But triangle OCD=triangle OEG
(OC=OE,DC=GE,<OCD=10=<OEG) ,soOD=DC=GO=GE.Therefore <ODB=<DOC+<DCO=
=10+10=20.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

2. Problem 1341 (sulotion 2)
Let K point on AB such that KA=KO , then <KAO=10=<KOA. I form the equilateral triangle
ΟΚL (the point L is located on the right of BC).Then the point K is the circumcenter the
triangle AOL.So <LAO=<LKO/2=60/2=30=<DAO so the point L belongs to AD.
Is triangle AOK=triangle COL(OA=OC,OK=OL,<KOA=10=<LOC),then LC=LO=KO=KA and
<LCO=10,so the point L belongs to BC.
Therefore the points D,L coincide and <ODB=<OLB=<LOC+<LCO=10+10=20.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

1. In order to prove "triangle AOK=triangle COL" above, how do we know that "10=<LOC" since we don't know yet that L is on BC?

3. https://ibb.co/df8rta
Let AO intersects BC in E. Since AB=BC and OA=OB we have DH is a perpendicular bisector of AE => FA=FE => EF=EB (5). From (3) and (4) and the AAS => triangle ADC=triangel ADF => AC=AF and DC=DF => AD is a perpendicular bisector of FC. Let the perpendicular bisector of BC intersects BC in M and AD in T => TC=TF=TB (6). From (5) and (6) => ET is a perpendicular bisector of FB, i.e. ET _|_ FB. Let TE intersects BF in K => triangle BEK~trianle TEM => D,O,E and T lie on a circle and from (7) => <ODE=<OTE=20.

4. Here is my solution...

1. How did u assume that the circle u drew is tangent to AD?

2. Intuition I guess. As I was looking and drawing the setup for the problem, I noticed the circle went through the apex of the large triangle and point D. And if you look closely at segment AD, that circle is NOT tangent to it. I don't know how else to explain it.

3. This is not proper geometric reasoning using accepted and time tested Ecludean principles and axioms

Sorry not acceptable

5. Extend BO to X so that AXC is an equilateral triangle.

OC bisects <BCX
So BO/BX = BC/XC = AB/AC =BD/DC since AD bisects <A

Hence OD//XC

So < BDO = < BCX = 20

Sumith Peiris
Moratuwa
Sri Lanka

1. Sumith - Will you please explain how BO/BX = BC/XC
Per Angle bisector theorem shouldn't it be BO/OX = BC/XC

2. @Sumith -- Typo correction reqd. BO/OX = BC/XC

3. Sorry for the typo

Presume All’s Well now

6. Problem 1341
Solution 2
From Problem 1274, circumradius of ∆ABC, R2 = a2b/(a+b)…(1)

Now since AD bisects < BAC, CD = ab/(a+b),
so CD.CB = a2b/(a+b) = R2 from (1).
Hence CO is tangential at O to circle BDO.
So < COD = 10 and hence < BDO = 10+10 = 20

Sumith Peiris
Moratuwa
Sri Lanka

7. Solution by Trigonometry:
Draw AE the height of T(ABC), then AE=EB=1 (valid for any dimension). Draw a perp. AF from O to AC, so AF=FC. Draw OC T(AOC) is isosc. and A(OCF)=10.
T(OCF) is sim. to T(AEC) now because FC=AC/2= 1/2sin(10) and and OF=AE=1/tan(10), the ratio of sim. is (1/2sin(10))/(1/tan10)=1/2cos10 and OF=1x1/2cos(10)= 1/2cos(10)
Per T(BDC) DC/sin(40)=2/sin(60) and DC=2sin(40)/sin(60)
Then, FD=FC-DC= 1/2sin(10)-2sin(40)/sin(60), finally tan(x)=OF/FD
tan(x)=(1/2cos(10))/(1/2sin(10)-2sin(40)/sin(60)) = 0.363970 and x=20

1. Sorry, I didn't follow the figure: I used A at the top and BC for the base of the triangle, the solution still valid.