## Monthly Archives: December 2014

### Basic Algebra for IITJEE Main

Main Problem: For an integer $n \geq 5$, let $S_{n}$ denote the sum of the products of the integers from 1 to n taken three at a time. Then, what is the value of $S_{10}$?

First Hint: Consider the terms in the expansion of $(1+2+3+ \ldots +10)^{3}$.

Second Hint: Classify these terms according to the number of times they occur.

Solution: There are totally 1000 terms in the expansion of

$(1+2+3+ \ldots +10)^{3}$. Terms of the form $ijk$ with i,j,k all distinct appear 6 times each. We are interested in the sum of these terms each such term considered only once. Then, there are terms of the form $i^{2}jk$ with

$i \neq j$. Each such product appears three times. Finally, there are terms of  the form $i^{3}$. Each such term appears once only. Hence,

$A=6S_{10}+3B+C$ equation I

where $A=(\sum_{i=1}^{10})^{3}$ and $C=\sum_{i=1}^{i=10}i^{3}$ and

$\sum_{i \neq j}^{10}i^{2}j$.

Using the well-known formulae given below:

$1+2+3+ \ldots +n = \frac {n(n+1)}{2}$ equation II

$1^{2}+2^{2}+3^{2}+ \ldots + n^{3}= \frac {n^{2}(n+1)^{2}}{4}$ equation III

A and C come out to be respectively $(55)^{3}$ and $55^{2}$.

To calculate B, note that if we add to B, products of the form $i^{2}i$, then we get all the terms of the product

$(1^{2}+2^{2}+3^{2}+ \ldots + n^{2})(1+2+3+\ldots +n)$.

The first factor equals $55 \times 7$ using the formula

$\sum n^{2}=\frac {n(n+1)(2n+1)}{6}$ equation IV

Hence, $B=6 \times (55)^{2}$. Putting these values in equation I, we get

$S_{10}=(1/6)(A-3B-C)=18150$.

More later

Nalin Pithwa

### Mathematics solves murder mystery !

Here is a puzzle for a young, mathematical Sherlock Holmes !

Question:

Suppose a mathematics professor was murdered while lying on his bed. The coroner gathered the following information:

At midnight, the professor’s body temperature was 94.6 degrees Fahrenheit. At 1 am, the body temperature was 93.4 degrees Fahrenheit. The room temperature was found to be 70 degrees Fahrenheit during this period. Assume a
reasonable body temperature for the professor before death, and use
Newton’s law of cooling to estimate when the professor expired.

More later Mr. Holmes of Mathematics 🙂

Nalin Pithwa

### Solutions of Triangles — a tricky IITJEE problem

Question (IITJEE 1978). Suppose $p_{1},p_{2},p_{3}$ are the altitudes through vertices A, B, C of a triangle ABC with area $\Delta$.

Prove that:

$\frac {1}{p_{1}} + \frac {1}{p_{2}} - \frac {1}{p_{3}}=\frac {2ab}{(a+b+c) \Delta} \cos^{2} (C/2)$

Proof:

The RHS looks daunting. But, if we bring the factor $\Delta$ in its denominator to the LHS, then the problem unfolds itself. Since, $\Delta =(1/2)ap_{1}=(1/2)bp_{2}=(1/2)cp_{3}$, the problem is equivalent to the following:

prove: $a+b-c=\frac {4ab}{a+b+c}\cos^{2}(C/2)$

If we write $2\cos^{2}(C/2)$ as $1+\cos C$, and multiply both the sides of by $a+b+c$, then the problem is reduced to  proving that

$(a+b)^{2}-c^{2}=2ab+2ab \cos C$.

which is the same as the cosine formula 🙂 🙂 🙂

More later…

Nalin Pithwa

### part 3 of 3 — Solutions to Pre RMO Oct 2014

Question Set A.

Question 20.

What is the number of ordered pairs $(A,B)$ where A and B are subsets of ${1,2,3,4,5}$ such that neither $A \subseteq B$ nor

$B \subseteq A$?

Solution. Just list down A and B explicitly. Note that A and B are disjoint and that $(A,B)$ is an ordered pair.

Question 13. For how many natural numbers n between 1 and 2014 (both inclusive) is $\frac {8n}{9999-n}$ an integer?

Solution. Firstly, note that $9999-n$ is even. Hence, n is odd.

Also, $8n \geq 9999-n$ to yield an exact integer, so $n \geq 1111$.

Now, do the one of the core tricks for problem solving in number theory. Plug and play with numbers 🙂

Put $n=1111$. This works.

Next, note that $(9999-n)k=8n$ for some positive integer k. Hence, we get

$9999k=(k+8)n$ From this we observe that $n=1111$ is the only possible solution. Hence, the answer is 1.

Note : Question 16: HW to be posted on the blog 🙂

More later,

Nalin Pithwa

### Part 2 of 3 — solutions to Pre RMO Oct 2014

Question Paper Set A:

11) For natural numbers x and y, let $(x,y)$ denote the greatest common divisor of x and y. How many pairs of natural numbers x and y with $x \leq y$ satisfy the equation $xy= x+y+ (x,y)$?

Solution:

Here, if the condition, $x \leq y$ were not there, the answer would be infinitely many. But, so, just substitute some small numbers and check what is happening. In fact, in any number theory problem, first we should play with small numbers in our head.

Upon substitution, you will find that only $(2,3), (2,4), (3,3)$ satisfy the equation. Hence, the answer is 3.

12) Let ABCD be a convex quadrilateral with

$\angle DAB = \angle BDC = 90 \deg$. Let the incircles of triangles ABD and BCD touch  BD at P and Q respectively, with P lying in between B and Q. If

$AD=999$ and $PQ=200$ then what is the sum of  the radii of the incircles of triangles ABD and BDC?

The main properties to  be used are Pythagoras’ theorem, that the angle bisectors of the vertices of a triangle meet at its incenter, the $\tan 2A$ formula of a triangle.

Let $C_{1}$ and $C_{2}$ be the incenter of $\Delta ABD$ and

$\Delta BCD$ respectively. Then, as shown in  the attached figure:

Since, $\Delta APD$ is a right angled isosceles triangle, $DP=DQ=999$, $\alpha 22.5 \deg$, so

$\tan 45 \deg = \frac {\tan 22.5 \deg}{1- \tan^{2} 22.5 \deg}= r_{1}/999$. Similarly, find the other radius and sum up the two.

Question 18.

Let f be a one-to-one function from the set of natural numbers to itself such that $f(mn)=f(m)f(n)$ for all natural numbers m and n. What is the least possible value of $f(999)$?

Probable Solution.

Use the Euler $\phi$ function. But, the answer this gives is different from the answer key.