HMMT 二月 2026 · 团队赛 · 第 3 题

3. [30] Let α and β be complex numbers such that αβ + α + β + 100 = 0 . Suppose that | α | = | β | = M for some nonnegative real number M . Determine, with proof, all possible values of M . Proposed by: Jason Mao Answer: 10 Solution 1: The answer is M = 10 , achievable by ( α, β ) = ( − 10 , 10) , for example. It remains to show that this is the only possible value. Let A and B be the points corresponding to α and β in the complex plane. The condition | α | = | β | = M implies that A and B lie on the circle ω centered at the origin O . ©2026 HMMT B A X O ′ A To translate the condition αβ + α + β + 100 , we let X be the point − 1 . Then we rewrite that condition to ( α + 1)( β + 1) = − 99 . ◦ In particular, the arguments of α + 1 and β + 1 sum to 180 . Therefore, AX and BX make the same angle to the x -axis. Moreover, since the right hand side − 99 is negative, we get that A and B lies on the same side with respect to OX . In particular, OX externally bisects ∠ AXB . Furthermore, AX · BX = 99 by comparing magnitudes. ′ Let A denote the reflection of A across OX . The angle condition in the previous paragraph implies ′ ′ A , X , B are collinear. Then A X · XB = 99 implies the power of X with respect to ω is − 99 . So 2 2 OX − M = − 99 , and since OX = 1 , it must be that M = 10 . Solution 2: As in the first solution, we show that 10 is the only possible value. B A T X O We use the geometric interpretation in the previous solution. In particular, let X be point − 1 , O be the origin, and A and B be points corresponding to α and β , respectively. Then, we have that OX externally bisects ∠ AXB and OA = OB = M , implying that AXOB is cyclic. Additionally, AX · BX = 99 . Let AB intersects OX at point T . Observe that ©2026 HMMT √ • Triangles XAT and XOB are similar, so XT · XO = XA · XB = 99 . (This is known as XA · XB inversion.) Since OX = 1 , we get that XT = 99 . 2 2 • Triangles OXA and OT B are similar, so OX · OT = OA = M . (This is known as “shooting 2 lemma”.) However, OX = 1 and OT = 100 , so M = 100 , implying that M = 10 . Solution 3: As in the first solution, we show that 10 is the only possible value. We rewrite the equation as 2 2 ( αβ + M ) + ( α + β ) = M − 100 . The crux of the problem is the following observation. Claim 1. We have that 2 αβ + M ∈ R . α + β Proof 1 (Geometric). Note that since | α | = | β | , we get that α + β lies on the internal angle bisector 2 2 of angle formed by points α , 0 , and β . Similarly, since | αβ | = M , we get that αβ + M lies on the internal angle bisector of angle formed by points 1 , 0 , and αβ . 2 αβ αβ + M α + β β α 2 0 M 2 However, those two angle bisectors are the same, so αβ + M and α + β have the same argument. 2 αβ + M Proof 2 (Algebraic). Let T = . We compare T against the conjugate T . From | α | = | β | = M , α + β 2 2 M M we get that α = and β = , so α β 4 M 2 2 2

• M αβ + M αβ + M αβ T = = = = T , 2 2 M M α + β α + β

α β so T ∈ R as desired. 2 Since αβ + M and α + β are real multiples of each other and they sum to a real number, we have two 2 possibilities: either M = 100 or α + β is real. The former implies M = ± 10 , but M > 0 , so M = 10 . Thus, we show that the latter is impossible. 2 Assume that α + β is real. Then αβ + M is real, and so αβ is real. Thus, we have two cases. • If α and β are both real , then since | α | = | β | = M , we have that α, β ∈ { M, − M } . Testing 2 2 2 all four possibilities gives αβ + α + β ∈ {− M , M − 2 M, M + 2 M } , each of these could not be − 100 . ©2026 HMMT 2 2 • If α and β are conjugates , then we have that αβ = M , so α + β = − M − 100 . Therefore, 2 M +100 Re α = − < − M , contradicting | α | = M . 2 Solution 4: As in the first solution, we show that 10 is the only possible value. 2 2 M M ¯ Clearly M ̸ = 0 , so α and β are both nonzero and thus have conjugates given by α ¯ = and β = . α β The conjugate of the relationship αβ + α + β + 100 = 0 then yields: 4 2 2 M M M

• • • 100 = 0 αβ α β 2 4 100 αβ + M ( α + β ) + M = 0 2 4 100 αβ − M ( αβ + 100) + M = 0 (since α + β = − αβ − 100 ) 2 4 2 (100 − M ) αβ + ( M − 100 M ) = 0 2 2 (100 − M )( αβ − M ) = 0 2 2 Thus, either M − 100 = 0 or αβ − M = 0 . The former implies M = ± 10 , but M > 0 , so M = 10 . Thus, we show that the latter is impossible. 2 Suppose that αβ = M . Then from | α | = | β | = M , we know that α and β are conjugates. Since 2 2 M +100 α + β = − ( αβ + 100) = − ( M + 100) , we get that Re α = − < − M , contradicting | α | = M . 2