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## Initiating a child into the world of Mathematical Science

“How do I involve my son in challenging mathematics? He gets good marks in school tests but I think he is smarter than school curriculum.”

“My daughter is in 4th grade. What competitions in mathematics and science can she participate? How do I help her to perform well in those competitions?”

“I have a 6 years old kid. He hates math. How do I change that?”

We often get queries and requests like these from parents around the world. Literally. In fact the first one came from Oregon, United States, second one from Cochin, India and last one from Singapore.

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## How do I involve my child in challenging mathematics?

“How do I involve my child in challenging mathematics? He gets good marks in school tests but I think he is smarter than school curriculum.”

“My daughter is in 4th grade. What competitions in mathematics and science can she participate in? How do I help her to perform well in those competitions?”

“I have a 6 years old kid. He hates math. How do I change that?”

We often get queries and requests like these from parents around the world. Literally. In fact, the first one came from Oregon, United States, the second one from Cochin, India, and the last one from Singapore.

Categories

## AMC 10 (2013) Solutions

12. In $(\triangle ABC, AB=AC=28)$ and BC=20. Points D,E, and F are on sides $(\overline{AB}, \overline{BC})$, and $(\overline{AC})$, respectively, such that $(\overline{DE})$ and $(\overline{EF})$ are parallel to $(\overline{AC})$ and $(\overline{AB})$, respectively. What is the perimeter of parallelogram ADEF?

$(\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad )$

Solution: Perimeter = 2(AD + AF). But AD = EF (since ABCD is a parallelogram).
Hence perimeter = 2(AF + EF).
Now ABC is isosceles (AB = AC = 28). Thus angle B = angle C. But EF is parallel to AB. Thus angle FEC = angle B which in turn is equal to angle C.
Hence triangle CEF is isosceles. Thus EF = CF.
Perimeter = 2(AF + EF) = 2(AF + EF) =2AC = $(2 \times 28)$ = 56.

Ans. (C) 56

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## USAJMO 2012 questions

1. Given a triangle ABC, let P and Q be the points on the segments AB and AC, respectively such that AP = AQ. Let S and R be distinct points on segment BC such that S lies between B and R, ∠BPS = ∠PRS, and ∠CQR = ∠QSR. Prove that P, Q, R and S are concyclic (in other words these four points lie on a circle).
2. Find all integers $(n \ge 3 )$ such that among any n positive real numbers $( a_1 , a_2 , ... , a_n )$ with $\displaystyle {\text(\max)(a_1 , a_2 , ... , a_n) \le n) (\min)(a_1 , a_2 , ... , a_n)}$ there exist three that are the side lengths of an acute triangle.
3. Let a, b, c be positive real numbers. Prove that $\displaystyle {(\frac{a^3 + 3 b^3}{5a + b} + \frac{b^3 + 3c^3}{5b +c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2))}$.
4. Let $(\alpha)$ be an irrational number with $(0 < \alpha < 1)$, and draw a circle in the plane whose circumference has length 1. Given any integer $(n \ge 3 )$, define a sequence of points $(P_1 , P_2 , ... , P_n )$ as follows. First select any point $(P_1)$ on the circle, and for $( 2 \le k \le n )$ define $(P_k)$ as the point on the circle for which the length of the arc $(P_{k-1} P_k)$ is $(\alpha)$, when travelling counterclockwise around the circle from $(P_{k-1} )$ to $(P_k)$. Suppose that $(P_a)$ and $(P_b)$ are the nearest adjacent points on either side of $(P_n)$. Prove that $(a+b \le n)$.
5. For distinct positive integers a, b < 2012, define f(a, b) to be the number of integers k with (1le k < 2012) such that the remainder when ak divided by 2012 is greater than that of bk divided by 2012. Let S be the minimum value of f(a, b), where a and b range over all pairs of distinct positive integers less than 2012. Determine S.
6. Let P be a point in the plane of triangle ABC, and $(\gamma)$ be a line passing through P. Let A’, B’, C’  be the points where reflections of the lines PA, PB, PC with respect to $(\gamma)$ intersect lines BC, AC, AB, respectively. Prove that A’, B’ and C’ are collinear.