Problem 1 of the Vietnam Mathematical Olympiad 1992

Let ABCD be a tetrahedron satisfying

  1. ∠ACD + ∠BCD = 180, and
  2. ∠BAC + ∠CAD + ∠DAB = ∠ABC + ∠CBD + ∠DBA = 180.

Find value of (ABC) + (BCD) + (CDA) + (DAB) if we know AC + CB = k and ∠ACB = α. Note: (Ω) denotes the area of shape Ω.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Mimi Nguyen of IBM)

Figure 1

Let [Φ] denote the plane containing shape Φ. Lay ΔABD, ΔACD and ΔBCD flat on the plane of ΔABC ([ABC]) as in Figure 1. The segments are dotted to show that they lie on the same plane [ABC]. Point D of triangle ACD is now at D; the same point D of triangle BCD is now at D, and that of triangle ABD is now at D.

We are already given ∠ACB = α. Now let ∠ACD = β, ∠BCD = χ, ∠DAB = μ, ∠BAC = ε, ∠CAD = δ, ∠CDA = γ, ∠ABD = λ, ∠ABC = φ, ∠CAD = ψ, AD = a, BD = b and CD = c (as shown in Figure 1).

Since ∠BAC + ∠CAD +∠DAB = ∠ABC +∠CBD + ∠DBA = 180 which can now be written as ε + δ + μ = φ + η + λ = 180, the three points D, A and D form a straight line, so do the three points D, B and D.

Also note that AD = AD = AD = a, BD = BD = BD = b, CD = CD = CD = c, and since A and B are the midpoints of DD and DD, respectively, AB || DD, and AB = DD.

Draw the altitude DK to the extension of DC. Since ∠ACD + ∠BCD (β + χ) = 180, or ∠ACD + ∠ACK = 180, we have ∠DCK = ∠ACB = α, and ∠DCD = 180 - α, and thus

(i)AB = DD = ccosα

Now lay flat the two triangles ACD and BCD to share the same side CD as shown in Figure 2.

Applying the law of the sine function for ΔABC,

  AB/sinα = AC/sinφ = BC/sinε = (AC + BC)/(sinφ + sinε)

or

(ii)AB = k sinα/(sinφ + sinε)

Equating the two values of AB in (i) and (ii), we have ccosα = ksinα/(sinφ + sinε) , or c(sinφ + sinε) = ksinα/cosα = 2ksinα, or

2csin(φ + ε) cos(φ - ε)= 2ksinα, or 2ccosα cos(φ - ε) = 2ksinα, or

(iii)ccos(φ - ε) = ktanα

Figure 2

Let S be the area (ABC) + (BCD) + (CDA) + (DAB); S = (DDD) (DCD) = 4(ABD) (DCD). But (ABD) = AB DH (H is the foot of D on AB). Now let DH = h, and using AB in (i), we have (ABD) = ccosα h .

On the other hand, (DCD) = DKc = csinα. Therefore, S = 2chcosα csinα = 2chcosα csinαcosα = ccosα (2h - csinα).

Now from C draw the altitudes CE and CF to AB and DD, respectively, and let CE = i, CF = j.

The above expression for S becomes S = c(2h j)cosα = c(2h j)cos∠CDD = c(2h j)[DD/(2c)]= (2h j)AB = (h + i)AB = 2(ACBD).

This result shows that S = 2[(ACD)+(BCD)] = 2(ABD) as shown in Figure 2.

Let H be the foot of D on AB and h = DH, we obtain S = h(AC + CB) = kh and now need to relate h to k and α.

Indeed, because AB || DD, we have

μ = γ + α/2, and λ = ψ + α/2, or

(iv)μ - λ = γ - ψ

And since ε + δ + μ = φ + η + λ = 180, it follows that

(v)φ - ε = δ + μ - λ - η

Substituting μ - λ from (iv) to (v) and note that ψ - η = β, we have

(vi)φ - ε = γ + δ - β

However, in ΔACD and ΔBCD because β + χ = 180, γ + δ = 180 - β, equation (vi) becomes

φ - ε = 180 - 2β or (φ - ε) = 90 - β, and equation (iii) becomes c cos(90 - β) = c sinβ = k tanα.

So now S = h(AC + CB) = kh = kcsinβ = ktanα.

This completes our analysis.