# VMO 1992 Problem 1

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, ∠D’’’AB = μ, ∠BAC = ε, ∠CAD’ = δ, ∠CD’A = 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 D’D’’’ and D”D’’’, respectively, AB || D’D”, and AB = ½ D’D’’.

Draw the altitude D”K to the extension of D’C. Since ∠ACD + ∠BCD (β + χ) = 180°, or ∠ACD’ + ∠ACK = 180°, we have ∠D”CK = ∠ACB = α, and ∠D’CD” = 180° - α, and thus

 (i) AB = ½D’D” = c×cos½α

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 c×cos½α = k×sinα/(sinφ + sinε) , or c×(sinφ + sinε) = k×sinα/cos½α = 2ksin½α, or

2c×sin½(φ + ε) cos½(φ - ε)= 2ksin½α, or 2c×cos½α cos½(φ - ε) = 2ksin½α, or

 (iii) c×cos½(φ - ε) = ktan½α

Figure 2

Let S be the area (ABC) + (BCD) + (CDA) + (DAB); S = (D’D”D’’’) – (D’CD”) = 4(ABD’’’) – (D’CD”). But (ABD’’’) = ½ AB × D’’’H (H is the foot of D’’’ on AB). Now let D’’’H = h, and using AB in (i), we have (ABD’’’) = ½c×cos½α× h .

On the other hand, (D’CD”) = ½ D”K×c = ½c²sinα. Therefore, S = 2ch×cos½α – ½c²sinα = 2ch×cos½α – c²sin½αcos½α = c×cos½α (2h - csin½α).

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

The above expression for S becomes S = c×(2h – j)×cos½α = c×(2h – j)×cos∠CD’D” = c×(2h – j)×[D’D”/(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 || D’D”, 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 ½(φ - ε) = 90° - β, and equation (iii) becomes c ×cos(90° - β) = c ×sinβ = k tan½α.

So now S = h’×(AC + CB) = kh’ = k×csinβ = k²tan½α.

This completes our analysis.