MathDB

1

Part of 2020 Dutch IMO TST

Problems(3)

<BAC = 2 <ABC wanted, AC + AI = BC given , incenter I

Source: 2020 Dutch IMO TST 1.1

11/21/2020
In acute-angled triangle ABC,IABC, I is the center of the inscribed circle and holds ∣AC∣+∣AI∣=∣BC∣| AC | + | AI | = | BC |. Prove that ∠BAC=2∠ABC\angle BAC = 2 \angle ABC.
geometryincenterequal angles
P_n (x) = x^{2n} + a_1x^{2n - 2} + a_2x^{2n - 4} +... + a_{n -1}x^2 + a_n

Source: 2020 Dutch IMO TST 2.1

11/22/2020
Given are real numbers a1,a2,...,a2020a_1, a_2,..., a_{2020}, not necessarily different. For every n≥2020n \ge 2020, define an+1a_{n + 1} as the smallest real zero of the polynomial Pn(x)=x2n+a1x2n−2+a2x2n−4+...+an−1x2+anP_n (x) = x^{2n} + a_1x^{2n - 2} + a_2x^{2n - 4} +... + a_{n -1}x^2 + a_n, if it exists. Assume that an+1a_{n + 1} exists for all n≥2020n \ge 2020. Prove that an+1≤ana_{n + 1} \le a_n for all n≥2021n \ge 2021.
algebrapolynomialinequalities
k = d (a) = d (b) = d (2a + 3b), no of positive divisors

Source: 2020 Dutch IMO TST 3.1

11/22/2020
For a positive number nn, we write d(n)d (n) for the number of positive divisors of nn. Determine all positive integers kk for which exist positive integers aa and bb with the property k=d(a)=d(b)=d(2a+3b)k = d (a) = d (b) = d (2a + 3b).
number theorynumber of divisors