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# 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
2 years, 4 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. $$\square$$

· 2 years, 4 months ago

I can't see why I mustn't call you god. · 2 years, 4 months ago

I can't see why I mustn't call you god. · 2 years, 4 months ago

I can't see why I mustn't call you god. · 2 years, 4 months ago

I can't see why I mustn't call you god. · 2 years, 4 months ago

I can't see why I mustn't call you god. · 1 year, 1 month ago

I can't see why I mustn't call you god · 2 years, 4 months ago

I can't see why I mustn't call you God. · 2 years, 4 months ago

I can't see why I mustn't call you god. · 2 years, 4 months ago

I can't see why I mustn't call you God. · 2 years, 4 months ago

I can't see why I mustn't call you god. · 2 years, 4 months ago

I can't see why I mustn't call you god. · 2 years, 4 months ago

I can't see why I mustn't call you god. · 2 years, 4 months ago

I can't see why I mustn't call you God. · 2 years, 4 months ago

Comment deleted Feb 06, 2016

I can't see why I mustn't call you god. · 1 year, 7 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 · 2 years, 3 months ago