Consider a triangle $ABC$ with $BC>AC$. The circle with center $C$ and radius $AC$ intersects the segment $BC$ in $D$. Let $I$ be the incenter of the triangle $ABC$ and $\Gamma$ be the circle that passes through $I$ and is tangent to the line $CA$ at $A$. The line $AB$ and $\Gamma$ intersect at a point $F$ with $F \neq A$. Prove that $BF=BD$.

Solution

Since the triangle $ACD$ is isosceles and $CI$ is the angle bisector the $I$ lies on the perpendicular bisector to the side $AD$. Therefore $AI=ID$. Since $\angle IAF = \angle CAI = \angle IFA$, then the triangle $AIF$ is isosceles and $AI=IF$. This implies that $IF=ID$ and the triangles $IDB$ and $IFB$ are congruent by SAS congruence. Therefore $BF=BD$ as desired.

Problem 2

Let $P(x)$ be a polynomial with real non-negative coefficients. Let $k$ be a positive integer and $x_1$, $x_2$, … , $x_k$ positive real numbers, such that $x_1 \cdot x_2 \cdot … \cdot x_k = 1$. Prove that

$$ P(x_1) + P(x_2) + … + P(x_k) \geq kP(1) $$

 

Solution

Let the polynomial $P(x)$ be $P(x) = \sum_{i=0}^{n} {a_i x^i}$. Therefore

$$ P(x_1) + P(x_2) + … + P(x_k) = \sum_{i=0}^{n} {a_i x_1^i} + \sum_{i=0}^{n} {a_i x_2^i} + … +\sum_{i=0}^{n} {a_i x_k^i} $$

The last expression is equal to

$$ \sum_{i=0}^{n} {a_i \left( x_1^i+x_2^i+…+x_k^i \right)} $$

Let us apply the AM-GM inequality to the numbers $x_1^i$, $x_2^i$, …, $x_k^i$:

$$ x_1^i+x_2^i+…+x_k^i \geq k \sqrt[k]{x_1^i \cdot x_2^i \cdot … \cdot x_k^i} = k $$

Therefore

$$ P(x_1) + P(x_2) + … + P(x_k) \geq \sum_{i=0}^{n} {a_i \cdot k} = k \sum_{i=0}^{n} {a_i} = k P(1) $$

Problem 3

A positive integer $N$ is said to be $interoceanic$ if its prime factorization

$$ N= x_1^{p_1} \cdot x_2^{p_2} \cdot … \cdot x_k^{p_k} $$

satisfies

$$ x_1+x_2+…+x_k = p_1+p_2+…+p_k $$

Find all interoceanic numbers less than $2020$.

Solution

Answer: $2^2$, $3^3$, $2^1 \cdot 3^4$, $2^2 \cdot 3^3$, $2^3 \cdot 3^2$, $2^4 \cdot 3^1$, $2^4 \cdot 5^3$, $2^5 \cdot 5^2$, $2^6 \cdot 5^1$, $2^8 \cdot 7^1$.

Notice that since $p_i \geq 2$, then for $x_1+x_2+…+x_k \geq 11$ we have

$$ N= x_1^{p_1} \cdot x_2^{p_2} \cdot … \cdot x_k^{p_k} \geq 2^{p_1} \cdot 2^{p_2} \cdot … \cdot 2^{p_k} = $$

$$2^{p_1+p_2+…+p_k} \geq 2^{11} > 2020$$

Therefore $x_1+x_2+…+x_k \leq 10$ and thus $p_1+p_2+…+p_k \leq 10$.

Notice that since $p_1 \geq 2$, $p_2 \geq 3$, $p_3\geq 5$, $p_4 \geq 7$, then for $k \geq 4$ we have

$$ p_1+p_2+…+p_k \geq 2+3+5+7 = 17 > 10 $$

Therefore $k \leq 3$ and there are no more than 3 different primes in the factorization of $N$.

Case 1: $k=1$. We have $x_1=p_1$ and therefore $N=p_1^{p_1}$. Notice that for the prime $p_1 \geq 5$ we will have $N = p_1^{p_1} \geq 5^5 >2020$. Therefore the only numbers that work are $2^2$ and $3^3$.

Case 2: $k=2$. We have $N=p_1^{x_1} p_2^{x_2}$. The only couples of distinct primes that satisfy the condition $p_1+p_2 \leq 10$ are $(2,3)$, $(2,5)$, $(2,7)$, $(3,5)$ and $(3,7)$.

For $N=2^{x_1}3^{x_2}$ we have $x_1+x_2=5$ and the numbers are $2^1 \cdot 3^4$, $2^2 \cdot 3^3$, $2^3 \cdot 3^2$, $2^4 \cdot 3^1$.

For $N=2^{x_1}5^{x_2}$ we have $x_1+x_2=7$ and the numbers are $2^4 \cdot 5^3$, $2^5 \cdot 5^2$, $2^6 \cdot 5^1$.

For $N=2^{x_1}7^{x_2}$ we have $x_1+x_2=9$ and the number is $2^8 \cdot 7^1$.

For $N=3^{x_1}5^{x_2}$ we have $x_1+x_2=8$. Since $N=3^{x_1}5^{x_2} \geq 3^7 \cdot 5^1 > 2020$, then there are no such numbers $N$.

For $N=3^{x_1}7^{x_2}$ we have $x_1+x_2=10$. Since $N=3^{x_1}7^{x_2} \geq 3^9 \cdot 7^1 > 2020$, then there are no such numbers $N$.

Case 3: $k=3$. We have $N=p_1^{x_1} p_2^{x_2}$. The only triple of distinct primes that satisfies the condition $p_1+p_2+p_3 \leq 10$ is $(2,3,5)$.

For $N=2^{x_1}3^{x_2}5^{x_3}$ we have $x_1+x_2+x_3=10$. Since $N=2^{x_1}3^{x_2}5^{x_3} \geq 2^8 \cdot 3^1\cdot 5^1 > 2020$, then there are no such numbers $N$.

Go to Day 1

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