Some random problems

 

I know, I know. Different font indeed. I have deleted a few of my MSE answers. I felt they weren't that good in quality. And a few questions are from my prev aops account which I have deactivated now.

I also have posted 10 IOQM types of problems. These can be used while preparing for IOQM.

Problem: Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$  are different positive real numbers. 

Proof: Note that by AM-GM $$\frac{ab}{c^3}+\frac{bc}{a^3}\ge \frac{2b}{ac}$$ and we also have $$\frac {b}{ac}+\frac{c}{ab}\ge \frac{2}{a}$$ .

Hence, $$\sum_{cyc}\frac{ab}{c^3}\ge\sum_{cyc}\frac{b}{ac}\ge\sum_{cyc}\frac{1}{a}$$

where everything we got is by applying AM-GM on $2$ terms and then dividing by $2$.

USA TST 2007: Triangle $ABC$ which is inscribed in circle $\omega$. The tangent lines to $\omega$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS$ is perpendicular to $AT$. Points $B_1$ and $C_1$ lies on ray $ST$ (with $C_1$ in between $B_1$ and $S$) such that $B_{1}T=BT=C_{1}T$.  Prove that the triangles $ABC$ and $AB_1C_1$ are similar to each other. 

Proof:

A beautiful problem indeed!

• Since $T$ is the point of intersection of tangent at $B$ and $C$ wrt  $\omega$ , we get that $BT=CT$. 
•  Now, using the conditions given us in the  problem, we get that $CT=B_{1}T=C_{1}T$. Hence $BCC_1B_1$ is cyclic quad.

• Now, define  $D$ as the midpoint of $BC$. Hence $TD\perp BC \implies TD \perp DS$ . 

• Since, $AS\perp AT$ ( given in the question ) , we get that $TDAS$ is cyclic.

Now, we are ready for angle chase! 

Let $\angle BAD=\angle TAC= \theta$ , $\angle BAC= \alpha$ , $\angle ABC= B$ .

Claim : $ACC_1S$ is cyclic

Proof : Note that by tangent- side theorem , we get that $\alpha= \angle CBT= \angle BCT \implies \angle DTC= 90-\alpha$ .

• Also, since $AS\perp AT$, we get that $\angle CAS= 90-\theta$ .

• Also, since $TDAS$ is cyclic, we get $$\angle DAT= \alpha -2\theta= \angle DST \implies \angle DTS = 90- (\alpha -2\theta) \implies \angle CTC_1=$$ $$2\theta \implies \angle TC_1C=90-\theta \implies \angle CC_1S=90+\theta$$ .

Hence we have $\angle CAS= 90-\theta$  and $\angle CC_1S=90+\theta$ . Hence $ACC_1S$ is cyclic.

Claim: $BB_1AS$ is cyclic 

Proof: Note that $\angle BAS= 90+ \alpha - \theta \implies \angle ASB= 90-( B+\alpha - \theta)$.

• By using the previous observation, that $\angle DST = \alpha -2\theta$ , we get that $\angle AST=\angle ASB_1= 90- (B+\theta)$ .

• Again, using the previous observations that , $\angle BTC= 180-2\alpha$ and $\angle CTC_1=2\theta$ , we get that $\angle BTB_1= 2 \alpha -2\theta \implies \angle B_1BT=90-(\alpha - \theta )$ .

Hence we get that $\angle B_1BA= 90+(B+\theta)$ .

• Now since, $\angle ASB_1= 90- (B+\theta)$ and  $\angle B_1BA= 90+(B+\theta)$ , we get that  $BB_1AS$ is cyclic .

Note that since $BB_1AS$ is cyclic and $ACC_1S$ is cyclic .Note that there is a spiral symmetry centered at $A$ dilating $\Delta ABC$ to $\Delta AB_1C_1$ . 

Hence  $\Delta ABC \sim \Delta AB_1C_1$. And we are done!

Australian MO: The right triangles $ABC$ and $AB_1C_1$ are similar and have opposite orientation. The right angles are at $C$ and $C_1$ and $\angle CAB = \angle C_1AB_1$. $M$ is the point of intersection of the lines $BC_1$ and $CB_1$. Prove that if the lines $AM$ and $CC_1$ exist, then they are perpendicular.

Proof: 

• Construction: Let $P$ is perpendicular from $A$ to $CB_1$, and $Q$ is perpendicular from $A$ to $BC_1$. 
• Now notice that $CQAB$ and $APC_1B_1$ are cyclic .
• So we have angle $\angle C_1PB_1 = \angle C_1AB_1 = \angle CQB = \angle CAB$. So $CQPE$ is cyclic and by POP we have $QM*MC_1 = CM*MP$.
•  So $M$  lies on the radical axis of circles around $CPA$ and $C_1QA$ .Also notice that $AM$ is this line. 
• As $AC$ and $AC_1$ are diameters of these circles we get that  $AM$ is perpendicular to line $RS$ where $R$ and $S$ are midpoints of $AC$ and $AC_1$. 

But  $RS$ is parallel to $CC_1$ . So we are  done

Russian MO: Given positive numbers $a_1, a_2, ..., a_m, b_1, b_2, ..., b_n$. Is known that $a_1+a_2+...+a_m=b_1+b_2+...+b_n$. 

Prove that you can fill an empty table with $m$ rows and $n$ columns with no more than $(m+n-1)$ positive number in such a way, that for all $i,j$ the sum of the numbers in the $i$-th row will equal to $a_i$, and the sum of the numbers in the $j$-th column -- to $b_j$.

Proof: we will use induction on $m+n$ . For $m+n= 2$, it is true .

assume it is true for $m+n=2,3....,k$

for $m+n=k$

put $X_{m,n} =min {a_m,b_n}$

Case1 : $X_{m,n} = a_m$

then strike off row m .

revise $b'_n = b_n - a_m$

then consider the new table with row sum $a_1,...a_{m-1}$ and column sums $b_1,b_2,...,b_{n-1}, b'_n$

by induction hypothesis the smaller table can still be filled with $k-1$ positive integers satisfying the question property . then these positive numbers with $X_{m,n}$ gives $K$ positive numbers which satisfy the conditions for bigger table ..

Case2: $X_{m,n}= b_n$

then strike off the column $n$.

the proof is similar.

so by induction, we are done.

AIME 1999: Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square.

Proof:

We check cases,

•      put $n=1,2,3,4,5,6,7,8,9,10.$ We get $n=1,9,10$ works.
•  $(n-10)^2=n^2-20n+100 < n^2-19n+99 < (n+10)^2$  we just have 19 cases
•   Equate them u are done ( I hope u don't die :))}

$$n^2 - 19n + 99 = k^2$$

$$n^2 - 19n + 99 - k^2 = 0$$

$D = 4k^2 - 396 + 361 = 4k^2 - 35$

now $4k^2 - 35 = m^2$

$(2k-m)(2k+m) = 35 \implies k = 9 or k = 3 \implies m = 17 or m = 1 $(ignore signs)

One gets $n = (19 \pm 1)/2$ or $n = (19\pm17)/2$

which gives $n= \boxed{10, 9, 1, 18}$

AIME 1999:For any positive integer $x$, let $S(x)$ be the sum of the digits of $x$, and let $T(x)$ be $|S(x+2)-S(x)|.$ For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values $T(x)$ do not exceed 1999?

Proof:

• So for unit digits being $0$ to $7,$ $T(X)$ is $2$
•  if not, the ten's place increases by 1 [ if ten's digit is $0,1,....., 8$]  then $T(x)$ is $7$
•    the ten's place increases by $1$ [ if ten's digit is $0,1,....., 8$]  then T(x) is 7
•    We get a pattern, it's $2,7,16,...,7+9k$
•   we get $223$ numbers

MPFG 2015: In how many different ways can 900 be expressed as the product of two (possibly equal) positive integers? Regard $m \cdot n$ and $n \cdot m$ as the same product.

Proof: There are $27$ divisors. So $(27-1)/2+1=14.$

MPFG 2013: When the binomial coefficient $\binom{125}{64}$ is written out in base 10, how many zeros are at the rightmost end?

Proof: find the number of $5$ powers and $2$ power. All the best. BTW it's $0.$ Powers of two cancel out.

$v_2(125!)={119}$

$v_2(64!)={63}$

$v_2(61!)={56}$

MPFG 2014: Four different positive integers less than 10 are chosen randomly. What is the probability that their sum is odd?

Proof:We have two cases i.e  3 odd 1 even or  1 odd 3 even.

$${5\choose 3} \cdot { 4\choose1} = 40$$

  $${4\choose 3}\cdot  {5\choose 1} = 20$$

 Hence $60.$ 

 The total is ${9\choose 4}=126$ hence $\frac{60}{126}.$ 

MPFG 2016: A permutation of a finite set $S$ is a one-to-one function from $S$ to $S$. A permutation $P$ of the set $\{ 1, 2, 3, 4, 5 \}$ is called a W-permutation if $P(1) > P(2) < P(3) > P(4) < P(5)$. A permutation of the set $\{1, 2, 3, 4, 5 \}$ is selected at random. Compute the probability that it is a W-permutation.

Proof: The all possible permutations of the set ${1, 2, 3, 4, 5 }$ is $5!$

We have $P(1) > P(2) < P(3) > P(4) < P(5).$

Note that $P(2),P(4)\in \{1,2,3\}.$

Take cases like $P(2)=1,P(4)=2$ etc (there will $6$ cases)

We get the probability as $\frac{2}{15}$

OMO spring 2019: Daniel chooses some distinct subsets of $\{1, \dots, 2019\}$ such that any two distinct subsets chosen are disjoint. Compute the maximum possible number of subsets he can choose.

Answer: 2020 :P

OMO 2020 spring: Given that the answer to this problem can be expressed as $a\cdot b\cdot c$, where $a$, $b$, and $c$ are pairwise relatively prime positive integers with $b=10$, compute $1000a+100b+10c$.

Proof: Answer is $\boxed{203010}$

We have $ac = 100a + c + 100$

So we get $(a-1)(c-100) = 200$

$a = 201$ and $c = 101.$

OMO 2017: A positive integer $n$ is called bad if it cannot be expressed as the product of two distinct positive integers greater than $1$. Find the number of bad positive integers less than $100.$

Proof: The bad numbers are $1$, primes, and the squares of primes. Compute it on your own :)

MPFG 2010: How many ordered triples of integers $(x, y, z)$ are there such that

$x^2 + y^2 + z^2 = 34?$

Proof: Bounding should work. So $6$ cases, take $x^2=0,1,4,9,16,25.$

So yes we are done! At least for this week! I am right now travelling to Odisha. I have boarding in like 10 minutes:P. comment down if you want to see some pics!

Also, how were the questions? I really don't know what level of IOQM problems I should send here. Hence I decided to send easy problems. If it's very easy, I will give fair-okayish level problems.

What do you want in the next post? I think it will be only IOQM problems since I am travelling, so write-ups are a bit hard. 

Stay safe and see you all soon,

Sunaina 💜