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 26th26^{th} digit from the left. Find it, given that 13 divides N.


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


5 ABCD is a cyclic quadrilateral with perpendicular diagonals AC and BD, intersecting at E. Prove that EA2+EB2+EC2+ED2=OA2+OB2+OC2+OD2EA^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=aAB=a and BC=bBC=b . Suppose r1r_1 is the radius of the circle passing through A and B and touching CD, and r2r_2 is the radius of the circle passing through B and C and touching AD. Show that r1+r25(a+b)8r_1+r_2 \geq \dfrac{5(a+b)}{8} .

Note by Satvik Golechha
4 years, 11 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

  1. At least 22 out of 55 balls of same color is same sized, so each color has at most 44 different sizes. Notice that one box must contain 65/2=33\ge \lceil 65/2\rceil=33 balls, and in that box there are 33/4=9\ge \lceil 33/4\rceil=9 balls of same color. Finally, among those 99 balls, there must be 9/4=3\ge \lceil 9/4\rceil=3 balls of same size. \square

  2. If the fold causes at the point MBCM\in BC, set BM=aBM=a and MC=bMC=b. Let NN be the midpoint of CDCD. Then CN=BC/2=(a+b)/2CN=BC/2=(a+b)/2 and a=BM=MNa=BM=MN by symmetry. So Pythagorean theorem on right MCN\triangle MCN gives MN2=MC2+CN2    a2=b2+(a+b)2/4    (a+b)(3a5b)=0MN^2=MC^2+CN^2\implies a^2=b^2+(a+b)^2/4\implies (a+b)(3a-5b)=0. Since a+b0a+b\neq 0 we have 3a5b=03a-5b=0 that is a/b=5/3a/b=5/3. \square

  3. Since 1311111113\mid 111111, the number found by removing first 6×4=246\times 4=24 ones, is divisible by 1313. Also we can replace the rightmost 6×4=246\times 4=24 ones by zeros. So if the unknown digit is dd then 131024(10+d)13\mid 10^{24}(10+d). By Fermat's little theorem 10121(mod13)10^{12}\equiv 1\pmod{13}, so 1310+d13\mid 10+d. It follows that d=3d=3. \square

  4. Let gcd(a,b)=g\gcd(a,b)=g, write a=gma=gm and b=gnb=gn with gcd(m,n)=1\gcd(m,n)=1. Now the equation rewrites to (m+n)/mn=g/c(m+n)/mn = g/c. Now gcd(g,c)=gcd(a,b,c)=1\gcd(g,c)=\gcd(a,b,c)=1 and gcd(m+n,mn)=gcd(m,n)=1\gcd(m+n,mn)=\gcd(m,n)=1 so m+n=gm+n=g. Substituting this value yields a+b=gm+gn=g(m+n)=g2a+b=gm+gn=g(m+n)=g^2. \square

  5. Notice that AOB=2ADB=2ADE=2(90DAE)=1802DAC=180COD\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 AOB=θ\angle AOB=\theta. Now (EA2+EB2)+(EC2+ED2)=AB2+CD2\left(EA^2+EB^2\right)+\left(EC^2+ED^2\right)=AB^2+CD^2 by Pythagorean theorem. Let RR be the radius. By law of cosine OA2+OB2=AB2+2R2cosθOA^2+OB^2=AB^2+2R^2\cos \theta and OC2+OD2=CD2+2R2cos(180θ)=CD22R2cosθOC^2+OD^2=CD^2+2R^2\cos\left(180^\circ-\theta\right)=CD^2-2R^2\cos\theta. Adding these we have OA2+OB2+OC2+OD2=AB2+CD2OA^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 r1=a2/8b+b/2r_1=a^2/8b+b/2 and r2=b2/8a+a/2r_2=b^2/8a+a/2. Subbing and simplifying reduces the inequality to a2/b+b2/aa+ba^2/b+b^2/a\ge a+b which follows by Titu's lemma. \square

Jubayer Nirjhor - 4 years, 11 months ago

Log in to reply

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

Krishna Ar - 4 years, 11 months ago

Log in to reply

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

Satvik Golechha - 4 years, 11 months ago

Log in to reply

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

Krishna Ar - 4 years, 11 months ago

Log in to reply

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

Satvik Golechha - 4 years, 11 months ago

Log in to reply

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

Abrar Nihar - 4 years, 11 months ago

Log in to reply

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

Mahdi Al-kawaz - 4 years, 11 months ago

Log in to reply

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

Satvik Golechha - 4 years, 11 months ago

Log in to reply

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

Kartik Sharma - 4 years, 11 months ago

Log in to reply

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

Satvik Golechha - 4 years, 11 months ago

Log in to reply

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

Kartik Sharma - 4 years, 11 months ago

Log in to reply

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

Harsh Shrivastava - 4 years, 11 months ago

Log in to reply

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

Harsh Shrivastava - 3 years, 8 months ago

Log in to reply

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

Satvik Golechha - 4 years, 11 months ago

Log in to reply

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 - 4 years, 9 months ago

Log in to reply

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

Aman Dubey - 3 years, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...