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 $r_1+r_2 \geq \dfrac{5(a+b)}{8}$ .

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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. $\square$
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 $a/b=5/3$ . $\square$
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 $d=3$ . $\square$
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 $a+b=gm+gn=g(m+n)=g^2$ . $\square$
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. $\square$
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. $\square$

Jubayer Nirjhor - 6 years, 7 months ago

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

Krishna Ar - 6 years, 7 months ago

Satvik Golechha - 6 years, 7 months ago

@Satvik Golechha – I can't see why I mustn't call you god.

@Krishna Ar – I can't see why I mustn't call you god.

Abrar Nihar - 6 years, 7 months ago

Mahdi Al-kawaz - 6 years, 7 months ago

@Mahdi Al-kawaz – I can't see why I mustn't call you god.

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

Kartik Sharma - 6 years, 7 months ago

@Satvik Golechha – I can't see why I mustn't call you God.

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

Harsh Shrivastava - 6 years, 6 months ago

Harsh Shrivastava - 5 years, 4 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

Curtis Clement - 6 years, 5 months ago

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

Aman Dubey - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$