IMO 1981 SL 2

Origin: BUL

Problem

A sphere S is tangent to the edges AB, BC, CD, DA of a tetrahe- dron ABCD at the points E, F, G, H respectively. The points E, F, G, H are the vertices of a square. Prove that if the sphere is tangent to the edge AC, then it is also tangent to the edge BD.

Solution

Lemma. Let E, F, G, H, I, and K be points on edges AB, BC, CD, DA, AC, and BD of a tetrahedron. Then there is a sphere that touches the edges at these points if and only if AE = AH = AI, BE = BF = BK, CF = CG = CI, DG = DH = DK. (∗) Proof. The “only if” side of the equivalence is obvious. We now assume (∗). Denote by ϵ, \varphi, \gamma, \eta, \iota, and \kappa planes through E, F, G, H, I, K perpendicular to AB, BC, CD, DA, AC and BD respectively. Since the three planes ϵ, \eta, and \iota are not mutually parallel, they intersect in a com- mon point O. Clearly, \triangleAEO ∼= A B C D H E F G \triangleAHO ∼= \triangleAIO; hence OE = OH = OI = r, and the sphere \sigma(O, r) is tangent to AB, AD, AC. To prove that \sigma is also tangent to BC, CD, BD it suﬃces to show that planes \varphi, \gamma, and \kappa also pass through O. Without loss of generality we can prove this for just \varphi. By the conditions for E, F, I, these are exactly the points of tangency of the incircle of \triangleABC and its sides, and if S is the incenter, then SE \perpAB, SF \perpBC, SI \perpAC. Hence ϵ, \iota, and \varphi all pass through S and are perpendicular to the plane ABC, and consequently all share the line l through S perpendicular to ABC.

Since l = ϵ \cap\iota, the point O will be situated on l, and hence \varphi will also contain O. This completes our proof of the lemma. Let AH = AE = x, BE = BF = y, CF = CG = z, and DG = DH = w. If the sphere is also tangent to AC at some point I, then AI = x and IC = z. Using the stated lemma it suﬃces to prove that if AC = x + z, then BD = y + w. Let EF = FG = GH = HI = t, \angleBAD = \alpha, \angleABC = \beta, \angleBCD = \gamma, and \angleADC = \delta. We get t2 = EH2 = AE2 + AH2 −2 \cdot AE \cdot AH cos \alpha = 2x2(1 −cos \alpha). We similarly conclude that t2 = 2y2(1−cos \beta) = 2z2(1−cos\gamma) = 2w2(1− cos \delta). Further, using that AB = x + y, BC = y + z, cos \beta = 1 −t2/2y2, we obtain AC2 = AB2 + BC2 −2AB \cdot BC cos \beta = (x −z)2 + t2 x y + 1 z y + 1 . Analogously, from the triangle ADC we get AC2 = (x −z)2 + t2(x/w + 1)(z/w + 1), which gives (x/y + 1)(z/y + 1) = (x/w + 1)(z/w + 1). Since f(s) = (x/s + 1)(z/s + 1) is a decreasing function in s, it follows that y = w; similarly x = z. Hence CF = CG = x and DG = DH = y. Hence AC \parallelEF and AC : t = AC : EF = AB : EB = (x + y) : y; i.e., AC = t(x + y)/y. Similarly, from the triangle ABD, we get that BD = t(x+y)/x. Hence if AC = x+z = 2x, it follows that 2x = t(x + y)/y ⇒2xy = t(x + y) ⇒BD = t(x + y)/x = 2y = y + w. This completes the proof. Second solution. Without loss of generality, assume that EF = 2. Con- sider the Cartesian system in which points O, E, F, G, H respectively have coordinates (0, 0, 0), (−1, −1, a), (1, −1, a), (1, 1, a), (−1, 1, a). Line AH is perpendicular to OH and AE is perpendicular to OE; hence from Pythagoras’s theorem AO2 = AH2 + HO2 = AE2 + EO2 = AE2 + HO2, which implies AH = AE. Therefore the y-coordinate of A is zero; analo- gously the x-coordinates of B and D and the y-coordinate of C are 0. Let A have coordinates (x0, 0, z1): then −\to EA(x0 + 1, 1, z1 −a) \perp−−\to EO(1, 1, −a), i.e., −\to EA \cdot −−\to EO = x0 + 2 + a(a −z1) = 0. Similarly, for B(0, y0, z2) we have y0 + 2 + a(a −z2) = 0. This gives us z1 = x0 + a2 + 2 a , z2 = y0 + a2 + 2 a . (1) We haven’t used yet that A(x0, 0, z1), E(−1, −1, a) and B(0, y0, z2) are collinear, so let A′, B′, E′ be the feet of perpendiculars from A, B, E to the plane xy. The line A′B′, given by y0x + x0y = x0y0, z = 0, contains the point E′(−1, −1, 0), from which we obtain (x0 + 1)(y0 + 1) = 1. (2)

In the same way, from the points B and C we get relations similar to (1) and (2) and conclude that C has the coordinates C(−x0, 0, z1). Similarly we get D(0, −y0, z2). The condition that AC is tangent to the sphere \sigma(O, OE) is equivalent to z1 = \sqrt a2 + 2, i.e., to x0 = a \sqrt a2 + 2 −(a2 + 2). But then (2) implies that y0 = −a \sqrt a2 + 2 −(a2 + 2) and z2 = − \sqrt a2 + 2, which means that the sphere \sigma is tangent to BD as well. This ﬁnishes the proof.