## Friday, February 10, 2012

### Problem 727: Triangle, Isosceles, Congruence, Angles, Measurement

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 727 details. 1. http://img197.imageshack.us/img197/8941/problem727.png

Let DD’ is the perpendicular bisector of AC and DD’ cut BC at M
AM is the angle bisector of angle BAC ( see picture)
Draw BK//AC and locate point L on BK such that BL=BA=BD
Note that m(BAM)=m(MAC)=44
Triangle ABL is isosceles => m(BLA)=1/2*(180-92)=44
So m(BAM)=m(BAL)=44 => A, M ,L are collinear
Triangle BKM congruence to LKM ( case ASA)
so BK=KL => triangle BDL is equilateral
and m(DBL)=60 => x=60-44=16

2. Nice proof, Peter. I applied Ceva's Trigonometric Theorem to Tr. ABC and landed up with: sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)
Can anyone tell how this is solved the answer, of course, being x=16 deg.?
Ajit

3. A trigonometric Proof:
B = 48°
∠ABD = 48° - x
ΔABD is isosceles
AD = 2 AB sin(∠ABD/2)= 2 c sin(24° - x/2)
∠BAD = ∠BDA = 66° + x/2
∠DAC = ∠DCA = 22° - x/2
ΔABD is isosceles
b = AC = 2 AD cos (22° - x/2)
b = 4c sin(24° - x/2) cos(22° - x/2)
sin B = 4 sin C sin(24° - x/2) cos(22° - x/2)
sin 48° = 4 sin 44° sin(24° - x/2) cos(22° - x/2)
sin 48° = (2 sin 44°)[2 sin(24° - x/2) cos(22° - x/2)]
sin 48° = (2 sin 44°)[sin(46° - x) + sin 2°]
cos 42° = 2 cos 46° sin(46° - x) + 2 sin 44°sin 2°
cos 42° = 2 cos 46° sin(46° - x) + cos 42° - cos 46°
sin(46° - x) = 1/2
46° - x = 30°
x = 16°

4. Dear Ajit:
sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)
=2sin(24-x/2)*cos(24-x/2)*sin(22+x/2)
Observe sin(66+x/2)and cos(24-x/2)get cancelled
since the angles are complementary
We now have
sin(x)= 2sin(24-x/2)*sin(22+x/2)= cos(2-x)-cos(46)
cos(46)=cos(2-x)-sin(x)=sin(88+x)-sin(x)=2cos(44+x)sin(44)
Again cos(46)and sin(44)get cancelled
since the angles are complimentary
So we are left with
1=2cos(44+x)
cos(44+x)= cos(60)
x=16 deg
Pravin

5. Thanks, Pravin.
My email: ajitathle@gmail.com
Ajit

6. Dear Ajit,

sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)

sin(x)*cos(24-x/2)=2sin(24-x/2)*cos(24-x/2)*sin(22+x/2), hence cos(24-x/2) get cancelled.

sin(x)=2sin(24-x/2)*sin(22+x/2). Using product to sum formula,

cos(90-x)=cos(2-x)-cos46

cos(2-x)-cos(90-x)=cos46. After using sum to product formula,

-2sin(46-x)*sin(-44)=sin44

-2sin(46-x)*[-sin44]=2sin30*sin44

2sin44*sin(46-x)=2sin30*sin44, hence 2sin44 get cancelled.

Finally sin(46-x)=sin30, then x=16.

7. Problem 727
Forms the isosceles trapezoid ABEC (the point Eis the right of point B and above the point C).
(ΒΕ//ΑC,<BAC=88=<ECA). Then AB=BD=EC=BE and BC is bisector <ACE.So ABEC is cyclic. But triangle ABD=triangle CED (AB=BD=CE, AD=DC,<BAD=<ECD).Then BD=DE=BE, so triangle BED is equilateral .Therefore x=<DBC=<DBE-<CBE=60-44=16.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

8. Solucion enviada por Pedro Miranda de Peru
Ver solucion aqui