31st IMO shortlist 1990/5 solution

31st IMO 1990 shortlist

Problem 5

The triangle ABC has unequal sides, centroid G, incenter I and orthocenter H. Show that angle GIH > 90^o.

Solution

Let N be the midpoint of OH. Then IN = (IO + IH)/2, so IH = 2IN - IO (we use bold to represent vectors). G lies on the line OH with OG = OH/3 (the Euler line), so OG = 2GN and hence IG = (2IN + IO)/3. Hence IH.IG = (4IN² - IO²)/3. We have the well-known results OI² = R² - 2rR (Euler's formula and IN = R/2 - r (Feuerbach's theorem - usually stated as "the incircle and the nine-point circle touch" - N is the center of the nine-point circle and R/2 is the radius of the nine-point circle).

Hence IH.IG = (R² - 4Rr + 4r² - R² + 2Rr)/3 = -2r(R - 2r)/3 < 0.

If you are fluent with vector formulae for the triangle, the following solution by Mehul Srivastav is straightforward

Use vectors origin O the circumcenter. Take the vector OA to be A etc. Then G = (A + B + C)/3, H = A + B + C (Euler line), I = (aA + bB + cC)/(a+b+c). The last formula is not so well-known, but is easy to verify. Check, for example, that b AI.AB = c AI.AC (for that it is convenient to relocate the origin to A).

We have to show that (G-I).(H-I) < 0, or G.H + I² - I.(G+H) < 0.

Note that since the origin is the circumcenter we have B.C = R²cos2A = R²cos²A - R²sin²A = (R² - a²/4) - a²/4 = R² - a²/2. Similarly for C.A and A.B. Obviously A² = B² = C² = R². Hence 3G.H = A² + B² + C² + 2A.B + 2B.C + 2C.A = 9R² - (a² + b² + c²).

We have I² = (aA + bB + cC)²/(a+b+c)² = ( (a² + b² + c²)R² + 2ab(R² - c²/2) + 2ac(R² - b²/2) + 2bc(R² - a²/2) )/(a+b+c)² = R² - (abc² + ab²c + a²bc)/(a+b+c)² = R² - abc/(a+b+c).

(3/4)(a+b+c)I.(G+H) = (aA + bB + cC).(A + B + C) = (a+b+c)R² + (a+b)A.B + (b+c)B.C + (c+a)C.A = 3(a+b+c)R² - (c²(a+b) + a²(b+c) + b²(a+c))/2.

So we wish to show that 3R² - (a²+b²+c²)/3 + R² -abc/(a+b+c) - 4R² + (2/3)(a²b + b²a + ... )/(a+b+c) < 0, or (a+b+c)(a²+b²+c²) + 3abc > 2(a²b + b²a + ... ) or a³ + b³ + c³ - (a²b + b²a + ... ) + 3abc > 0 or a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) > 0 (*).

wlog a > b > c. So a(a-c) - b(b-c) > 0. Hence a(a-b)(a-c) + b(b-a)(b-c) > 0. Obviously c(c-a)(c-b) > 0. So (*) holds and hence the result.

31st IMO shortlist 1990