- angle BCD equals 50 degrees.
- angle BCD equals 70 degrees.

Source: rec.puzzles.

Solutions were received from a few people. Denis Borris found he could actually solve the problem analytically with use of the Sine Law and Cosine Law. Radu Ionescu found the solutions purely through comparisons of angles. And Philippe Fondanaiche sent a solution using similar triangles.

- For BCD=50, CDE is 80 degrees.
- For BCD=70, CDE is 20 degrees.

Philippe Fondanaiche's solutions:

Question 1:angle BCD = 50° Let BC = a. As ABC is isosceles with AB = AC, angle DBC = (180 - 20)/2 = 80° Let F on AC such as angle CBF = 20°.BCF is isosceles as ABC with BF = BC = a. As angle BDC = 180 - angle DBC - angle BCD = 180 - 80 - 50 = 50°,then BCD is isosceles with BD = BC = a. As angle DBF = 80 - angle CBF = 80 - 20 = 60°,DBF is equilateral ==> DF = a Moreover angle EBF = angle CBE - angle CBF = 60 - 20 = 40° and angle BEF = 180° - angle CBE - angle BCE = 180 - 60 - 80 = 40°. Therefore FBE is isosceles with FE = FB = a. Then as DF = EF = a,FDE is isosceles with angle DFE = 180 - angle BFC - angle BFD = 40°. So angle EDF = (180 - 40)/2 = 70° and angle CDE = angle EDF + angle CDF = 70 + (60 - 50) = 80° As a conclusion, angle CDE = 80° Question 2:angle BCD = 70° Let F on BA such as BF = BC = a. It is easy to check with the identities mentioned in the question 1(F is the equivalent of the point D),that in the triangle BFE,we have :angle EBF = 20°,angle BFE = 130° and angle BEF = 30° and that in the triangle CFD, we have:angle CDF = 180 - angle BCD - angle CBD = 180 - 70 - 80 = 30°, angle DFC = 180 - angle BFC = 180 - 50 = 130° and angle DCF = angle BCD - angle BCF = 70 - 50 = 20°. So the triangles BFE and CFD having the same angles,are similar.Then DF/EF=CF/BF. As angle BFC = angle DFE = 50°, the triangles EDF and BCF are similar. As BCF is isosceles with BF = BC,then EDF is isosceles too with ED = EF. So angle EDF = 50° and angle CDE = 50 - 30 = 20° As a conclusion,angle CDE = 20°

Radu Ionescu's solutions:

1) Let F on AB , EF paralel to BC and O intersection of BE and CF ==> FEO and OBC are triangle Equilateral angle(BCD)=angle(BDC)=50 ==> BD=BC =BO ==> BDO triangle isoscele ==>angle(BOD)=80 ==> angle(DOF)=40 =angle(OFD) ==>DF=DO DF=DO, FE=OE and DE=DE==> triangle(DFE)= triangle(DOE)==> angle(FDE)=angle(ODE)=50 (a) angle(OEC)=angle(ODB)-angle(CDB)=80-50=30 (b) (a) and (b)==>angle(CDE)=80 2) Let F on AB and angle(FCB)=50 ==>angle(CFE)=80 [ see 1) ] O intersection of BE and CD. In FOCB angle(FBO)=angle(FCO)=20 ==>FOCB is a cyclic quadrilateral (a) H intersection of FC and BE. In triangle FHE angle(HFE)=80=angle(CFE), angle(FHE)=70==> angle(FEH)=30=angle(FEO) In FOED angle(FDO)=30=angle(FEO)==>FOED is a cyclic quadrilateral (b) (a) ==> angle(OFC)=angle(OBC)=60 and angle(CFE)=80 ==>angle(OFE)=20 (b)==>angle(ODE)=angle(OFE)=20 ==> angle(CDE)=20

Denis Borris' solutions:

A E D C B (shoot! AE is really a bit longer than EC, but what the heck...) typing reducers: tABC = triangle, aABC = angle, s30 = sin(30), [Assume AB=AC=1] STEP 1: BC = s20 / s80 = w [Sine Law] since tBCD is isosceles then BD = w STEP 2: CD = s20 / s50 = x [Sine Law] STEP 3: using tBCE: CE = w s60 / s40 = y Step 4: using tCDE and cosine law: DE = sqrt(x^2 + y^2 - 2xy c30) = z Step 5: (we now have lengths of tCDE's 3 sides, along with aDCE = 30) so: sin(aCDE) = (y s30) / z = .9848077530122080588... = 80 (eighty!) degrees. ...oh oh: just heard P de Fermat's bones play the drums against the sides of his casket!! [And a similar answer for the second problem. - KD]

Update July 11, 2006. Ron L.J. van den Burg felt a little more rigor could be applied:

I just happened
to see the December 1, 1998 problem and saw that "analytical" solution was
not *really* analytical. It uses the approximation ".9848077530122080588...
= 80 (eighty!) degrees"

If you are
interested, I found an analytical solution with pure Sine, Cosine and
Tangent Laws only.

- Let B
have coordinate (-1, 0) and let C have coordinate (1, 0). A, then lies
on the y-axis.

- Let
P have coordinate (Px, Py) be a point on AB such that angle BCP equals
50 degrees.

- Let Q
have coordinate (Qx, Qy) be a point on AC such that angle CBQ equals 60
degrees.

- Abbreviate Cosine(10 degrees) with c10, Sine(70 degrees) with s70, Tangent(40 degrees) with
t40 etc.

Step 1: use
basic high school math with lines and intersections to determine the
coordinates of P and Q.

- So, the
line trough AB reads f(x) = c80 x + c80. The line trough CP reads g(x) =
- c50 x + c50.

- Then, you
find Px = (t50 - t80) / (t50 + t80) and Qx = (t80 - t60) / (t80 + t60).

- Filling
in Px (and Qx) in f(x) or g(x) then gives Py = 2 t50 t80 / (t50 +
t80) and Qy = 2 t80 t60 / (t80 + t60).

If we
define point R as the intersection of the lines through BC and PQ, then
let's find the angle PRB: just the Als we nu de gezochte lijn door P en
Q helemaal doortrekken tot waar hij de lijn door de basis snijdt, zeg
onder hoek alpha, dan is het gevraagde bewijs equivalent met de vraag om
te bewijzen dat alpha 30 graden is. Oftewel: bewijs dat de
richtingscoefficient van de lijn door P en Q: (Qy - Py) / (Qx - Px)
gelijk is aan t30.

The slope /
gradient / tangent / angle of inclination of the line trough PQ is then
: 2 s80 s80 s10 / (s20 s50 + s30 s40).

Step 2: use
the Sine / Cosine Laws for sin(a+b) en cos(a+b) taught at high school.

- c30 = c20
c10 - s20 s10 = ... = 4 c10 c10 c10 - 3 c10.

- So 2 c10
c30 = 8 c10 c10 c10 c10 - 6 c10 c10 = 2 c20 c20 + c20 - 1 (check this
result by using c20 = 2 c10 c10 - 1.)

- Multiply
both sides of the equation with 2 s10 c10 (left) and s20 / 2
(right). This is allowed because 2 s10 c10 = s20 / 2 and is not equal to
zero.

- So 2 c10
c10 s10 c30 = s20 c20 c20 + s20 c20 / 2 - s20 / 2 = s20 c40 / 2 + s40 /
4.

- Now
use the equalities c40 = s50 and s30 = 1/2.

- This results to: 2 c10 c10 s10 c30 =
s20 s50 s30 + s40 s30 s30 = (s20 s50 + s40 s30) s30.

- Now use the equality: c10 = s80.

- Then: 2 s80 s80 s10 c30 = (s20 s50 +
s40 s30) s30.

In the end:

- 2 s80 s80 s10 / (s20 s50 + s40 s30) =
s30 / c30 = t30.

Step 3: Now
combine steps 1 and 2: the slope /gradient of the line through PQ is 30
degrees. The rest is trivial: PRB = 30 degrees => (using RBQ = 180 - 60
= 120) => RQB = 30 degrees => PQC = 70 degrees => (using PCQ = 80 - 50 =
30) => CPQ = 80 degrees.

Kind
regards,

Ron L.J. van den Burg

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