## Thursday, December 1, 2011

### Problem 696: Triangle, Angles, Auxiliary Construction, Congruence, Mind Map

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

Click the figure below to see the complete problem 696.

1. Any ideas on this solution? I am breaking my head with this particular problem. Thanks

2. Sorry I don't have a proof, but I messed around a lot with the law of sines between triangles ABD and BDC, along with trigonometric laws, and the use of my calculator, but I did get an answer. Have fun making a proof:

x= 10, therefore m ABD= 30

3. I've made half of the progress. With a perpendicular to AC through B intersecting CD in N we have the triangle ANB. The angle DAN=3x because ANC is isosceles. Now, AD is the bisector of BAN. If we prove that ND bisects ANB as well, we can conclude that BD bisects ABN so we can calculate x=10.
But, I can't prove that ND bisects BNA...

4. I propose a solution to problem 696, due to Daniel Kawai, brazilian student.
ABC is isosceles with AB = CB. Besides,
ADB = 180º - 3x, BDC = 11x,
and DBC = 180º - 15x > 0,
so 0 < x < 180º/15 = 12º.
Let’s use sine rule.
In tr. ABD, AD/AB = sin x/sin 3x.
In tr. CBD, CB/CD = sin 11x/sin 15x.
But in ACD, AD/CD = sin 3x/sin 5x, thus
(sin x.sin 11x)/(sin 3x.sin 15x) = sin 3x/sin 5x,
and sin x.sin 5x.sin 11x = (sin 3x)^2.sin 15x.
From this equality we get, successively:
(cos 4x – cos 6x).sin 11x = (1 – cos 6x).sin 15x
sin 11x.cos 4x – sin 11x.cos 6x = sin 15x – sin 15x.cos 6x
sin 15x + sin 7x – sin 17x – sin 5x = 2sin 15x – sin 21x – sin 9x
sin 21x – sin 5x – sin 17x – sin 15x + sin 9x + sin 7x = 0
2sin 8x.cos 13x – 2sin 16x.cos x + 2sin 8x.cos x = 0
2sin 8x.cos 13x – 4sin 8x.cos 8x.cos x + 2sin 8x.cos x = 0
cos 13x – 2cos 8x.cos x + cos x = 0
cos 13x – cos 9x – cos 7x + cos x = 0
-2sin 11x.sin 2x + 2sin 4x.sin 3x = 0
-2sin 11x.sin 2x + 4sin 2x.cos 2x.sin 3x = 0
sin 11x - 2cos 2x.sin 3x = 0
sin 11x – sin x – sin 5x = 0
2sin 5x.cos 6x – sin 5x = 0
Finally cos 6x = 1/2, 6x = 60º and x = 10º.

5. Here is a new idea for proble 696.
Let be BM the altitude and median of the triangle ABC, O the intersection of BM and the extension of AD, and G the point on BM such that ang(GAM) = 3x. Thus ang(AGD) = 6x (external angle of AGC) and ang(AGO) = 90º + 3x (external angle of triangle AGM). Let be y = ang(DBO) and z = ang(DGO), then x+y = 90º - ang(BAM) = 90º - 7x and y = 90º - 8x. Let be z = ang(AGO) – ang(AGD) = 90º + 3x – 6x = 90º - 3x.
Since y = 90º - 8x > 0, then 0 < 8x < 90º. Let’s apply the sine’s rule to triangles ADG and ODG. We get AD/AG = sin 6x/sin (ang ADG) and OD/OG = sin z/sin (ang ODG).
But sin (ang ADG) = sin (ang ODG), so
AD/OD = (AG.sin 6x)/(OG.sin z) = (AG.sin 6x)/(OG.cos 3x) (1).
Let’s apply the sine’s rule to triangles ADB and ODB. We get AD/AB = sin x/sin (ang ADB) and OD/OB = sin y/sin (ang ODB).
But sin (ang ADB) = sin (ang ODB), so
AD/OD = (AB.sin x)/(OB.sin y) = (AB.sin x)/(OB.cos 8x) (2).
Since AO is the angle bisector of ang(BAG), then AG/OG = AB/OB. Comparing (1) and (2), we get sin 6x/cos 3x = sin x/cos 8x. From this trigonometric equation we obtain, successively, sin 6x.cos 8x = sin x.cos 3x, then sin 14x – sin 2x = sin 4x – sin 2x, then sin 14x = sin 4x. The hypothesis 14x = 4x is impossible. Hence 14x + 4x = 180º and x = 10º.

6. Problem 696 revisited:
Let ∠DBC = y
BD/sin2x=AB/sin3x by sine rule in ∆ABD
BD/sin4x=BC/sin(4x+y) by sine rule in ∆ACD
sin4x/sin2x=sin(4x+y)/sin3x since AB=BC
sin(4x+y).sin2x=sin4x.sin3x
15x+y=180° since A+B+C=180°
4x+y=180°-11x
sin11x.sin2x=sin4x.sin3x
sin11x=2sin3x.cos2x=sin5x+sinx
sin11x - sinx = sin5x
2cos6x.sin5x= sin5x
cos6x=1/2
6x=60°
x=10°
{note:sin5x=0 gives x=36°,A=C=252°absurd)

7. Here is a solution for problem 696 without trigonometry.
Let E be a point in the exterior of ABC such that ang(BCE) = ang(BAD) = 2x, and ang(CBE) = ang(ABD) = x. Let F be the intersection of AD and the altitude BH. Let’s write ang(DBF) = y.

1) Triangles ABD and CBE are congruent (ASA) so BD = BE. Furthermore ang(DBE) = ang(DBC) + ang(CBE) = ang(DBC) + x = ang(DBC) + ang(ABD) = ang(ABC) so triangles DBE and ABC are similar. Hence ang(BDE) = ang(BED) = 7x.

2) Since ang(BDF) = 3x (external angle) then ang(FDE) = 4x.

3) Since ang(CDF) = 8x (external angle) then ang(CDE) = 4x.

4) Since ang(FCA)= ang(FAC) = 5x, and ang(DCA) = 3x, then ang(DCF) = 2x. Also ang(FCB) = 2x. So ang(FCE) = 4x.

5) Since ang(FDE)= ang(FCE) = 4x, quadrilateral DFEC is cyclic. Let be G the intersection of BC and the circumcircle of DFEC. We have ang(DCG) = ang(DEG) = 4x, so ang(BEG) = 3x.

6) As we’ve seen, ang(DCF) = ang(FCG) = ang(GCE) = 2x, so DF = FG = GE.

From the results above we conclude that triangles BDF and BEG are congruent (SAS). Hence y = x. In triangle BAH, 7x + 2x = 90º, so x = 10º

8. BH, bisector of AC, intersect CD in E.Let be AP the extension of AE such tha AP = BA = BC.Triangule DAP congruent to DAB (SAS) so DPE = x. It is not hard to see that BDP = 6x then DPB = 90º-3x and HEC = DEB = 90-3x hence quadrilateral BDEP is cyclic then DBE = DPE = x. Now is easy to see that x =10º.I challenge you to solve the problem 199.

9. To Anonymous (August 2, 2012)
Greetings.
The solution of problem 696 that you proposed is the simplest that I have seen, but there are some conclusions that I didn’t quite understand.
1) DAP and DAB are congruent by SAS? Shouldn’t the angle be between the two sides?