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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## Comments

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TopNewestIf 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. \(\square\)

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

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

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

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

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

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

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

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

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

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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

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

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