# Olympiad Practice Problems Part 2

After Part 1, here's the second one. This time the problems are from RMO, conducted in various states of India. Have fun, and please post hints/solutions as comments, like, re-share, and enjoy!

1 ** Two boxes contain between them total 65 balls of several different sizes. Each ball is of any one colour from white, black, red, or yellow. If we take any 5 balls of the same colour, at least two of them always have same size. Prove that there are at least 3 balls which have the same box, same colour, and the same size.

2 A square sheet of paper ABCD is so folded that B falls on the mid-point of CD. Prove that the crease will divide BC in the ratio 5:3.

3 'N' is a 50 digit number. All digits of N are 1, except the $26^{th}$ digit from the left. Find it, given that 13 divides N.

4 If $a,b,c \in \mathbb{Z^+}$, and $\gcd(a,b,c)=1$, and $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, prove that $(a+b)$ is a square.

5 ABCD is a cyclic quadrilateral with perpendicular diagonals AC and BD, intersecting at E. Prove that $EA^2+EB^2+EC^2+ED^2=OA^2+OB^2+OC^2+OD^2$, where O is the center of the circle.

6 Let ABCD be a rectangle with $AB=a$ and $BC=b$ . Suppose $r_1$ is the radius of the circle passing through A and B and touching CD, and $r_2$ is the radius of the circle passing through B and C and touching AD. Show that $r_1+r_2 \geq \dfrac{5(a+b)}{8}$ .

Note by Satvik Golechha
4 years, 11 months ago

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1. At least $2$ out of $5$ balls of same color is same sized, so each color has at most $4$ different sizes. Notice that one box must contain $\ge \lceil 65/2\rceil=33$ balls, and in that box there are $\ge \lceil 33/4\rceil=9$ balls of same color. Finally, among those $9$ balls, there must be $\ge \lceil 9/4\rceil=3$ balls of same size. $\square$

2. If the fold causes at the point $M\in BC$, set $BM=a$ and $MC=b$. Let $N$ be the midpoint of $CD$. Then $CN=BC/2=(a+b)/2$ and $a=BM=MN$ by symmetry. So Pythagorean theorem on right $\triangle MCN$ gives $MN^2=MC^2+CN^2\implies a^2=b^2+(a+b)^2/4\implies (a+b)(3a-5b)=0$. Since $a+b\neq 0$ we have $3a-5b=0$ that is $a/b=5/3$. $\square$

3. Since $13\mid 111111$, the number found by removing first $6\times 4=24$ ones, is divisible by $13$. Also we can replace the rightmost $6\times 4=24$ ones by zeros. So if the unknown digit is $d$ then $13\mid 10^{24}(10+d)$. By Fermat's little theorem $10^{12}\equiv 1\pmod{13}$, so $13\mid 10+d$. It follows that $d=3$. $\square$

4. Let $\gcd(a,b)=g$, write $a=gm$ and $b=gn$ with $\gcd(m,n)=1$. Now the equation rewrites to $(m+n)/mn = g/c$. Now $\gcd(g,c)=\gcd(a,b,c)=1$ and $\gcd(m+n,mn)=\gcd(m,n)=1$ so $m+n=g$. Substituting this value yields $a+b=gm+gn=g(m+n)=g^2$. $\square$

5. Notice that $\angle AOB=2\angle ADB=2\angle ADE=2\left(90^\circ-\angle DAE\right)=180^\circ-2\angle DAC=180^\circ-\angle COD$. Now let $\angle AOB=\theta$. Now $\left(EA^2+EB^2\right)+\left(EC^2+ED^2\right)=AB^2+CD^2$ by Pythagorean theorem. Let $R$ be the radius. By law of cosine $OA^2+OB^2=AB^2+2R^2\cos \theta$ and $OC^2+OD^2=CD^2+2R^2\cos\left(180^\circ-\theta\right)=CD^2-2R^2\cos\theta$. Adding these we have $OA^2+OB^2+OC^2+OD^2=AB^2+CD^2$ as well. $\square$

6. Clearly the circles touch at the midpoints of the sides. It is then a straight computation using cosine law to find that $r_1=a^2/8b+b/2$ and $r_2=b^2/8a+a/2$. Subbing and simplifying reduces the inequality to $a^2/b+b^2/a\ge a+b$ which follows by Titu's lemma. $\square$

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 4 years, 11 months ago

I can't see why I mustn't call you God.

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 4 years, 11 months ago

I can't see why I mustn't call you God.

- 4 years, 11 months ago

I can't see why I mustn't call you god

- 4 years, 11 months ago

I can't see why I mustn't call you god.

- 3 years, 8 months ago

I can't see why I mustn't call you God.

- 4 years, 11 months ago

I can't help but imagine people attempting Q2 in the exam and folding their paper, leaving the invigilators thinking 'what sort of exam is this?" :P

- 4 years, 9 months ago

Hehehe my mom too asked me what kind of math is this

- 3 years, 8 months ago

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