MathDB

P2

Part of 2023 Israel TST

Problems(5)

Equal areas in a pyramid

Source: 2023 Israel TST Test 1 P2

3/23/2023
Let SABCDESABCDE be a pyramid whose base ABCDEABCDE is a regular pentagon and whose other faces are acute triangles. The altitudes from SS to the base sides dissect them into ten triangles, colored red and blue alternatingly. Prove that the sum of the squared areas of the red triangles is equal to the sum of the squared areas of the blue triangles.
TSTgeometryareas3D geometrypyramid
Snakes in an 8 x 8 chessboard

Source: 2023 Israel TST Test 2 P2

3/23/2023
In an 8×88 \times 8 grid of squares, each square was colored black or white so that no 2×22\times 2 square has all its squares in the same color. A sequence of distinct squares x1,…,xmx_1,\dots, x_m is called a snake of length mm if for each 1≤i<m1\leq i <m the squares xi,xi+1x_i, x_{i+1} are adjacent and are of different colors. What is the maximum mm for which there must exist a snake of length mm?
TSTcombinatoricsColoring
A(n) divides B(n)

Source: 2023 Israel TST Test 3 P2

3/23/2023
For each positive integer nn, define A(n)A(n) to be the sum of its divisors, and B(n)B(n) to be the sum of products of pairs of its divisors. For example, A(10)=1+2+5+10=18A(10)=1+2+5+10=18 B(10)=1⋅2+1⋅5+1⋅10+2⋅5+2⋅10+5⋅10=97B(10)=1\cdot 2+1\cdot 5+1\cdot 10+2\cdot 5+2\cdot 10+5\cdot 10=97 Find all positive integers nn for which A(n)A(n) divides B(n)B(n).
TSTnumber theoryDivisors
Concyclicity with symmedian

Source: 2023 Israel TST Test 5 P2

3/23/2023
Let ABCABC be an isosceles triangle, AB=ACAB=AC inscribed in a circle ω\omega. The BB-symmedian intersects ω\omega again at DD. The circle through C,DC,D and tangent to BCBC and the circle through A,DA,D and tangent to CDCD intersect at points D,XD,X. The incenter of ABCABC is denoted II. Prove that B,C,I,XB,C,I,X are concyclic.
geometryTSTsymmedianincenter
Distinct partial sums

Source: 2023 Israel TST Test 7 P2

5/9/2023
Let n>3n>3 be an integer. Integers a1,…,ana_1, \dots, a_n are given so that ak∈{k,−k}a_k\in \{k, -k\} for all 1≤k≤n1\leq k\leq n. Prove that there is a sequence of indices 1≤k1,k2,…,kn≤n1\leq k_1, k_2, \dots, k_n\leq n, not necessarily distinct, for which the sums ak1a_{k_1} ak1+ak2a_{k_1}+a_{k_2} ak1+ak2+ak3a_{k_1}+a_{k_2}+a_{k_3} ⋮\vdots ak1+ak2+⋯+akna_{k_1}+a_{k_2}+\cdots+a_{k_n} have distinct residues modulo 2n+12n+1, and so that the last one is divisible by 2n+12n+1.
TSTcombinatoricsmodular arithmeticabstract algebra