## Tag Archives: ISI entrance

### Pre-RMO — training

Problem 1:

If a, b, c, and d satisfy the equations $a+7b+3c+5d=0$ $8a+4b+6c+2d=-16$ $2a+6b+4c+8d=16$ $5a+3b+7c+d=-16$

then what is the numerical value of $(a+d)(b+c)$?

Problem 2:

Suppose x and y are positive integers with $x>y$ and $3x+2y$ and $2x+3y$ when divided by 5, leave remainders 2 and 3, respectively. It follows that when $x-y$ is divided by 5, the remainder is necessarily equal to

(A) 2

(B) 1

(C) 4

(D) none of the foregoing numbers

Problem 3:

The number of different solutions $(x,y,z)$ of  the equation $x+y+z=10$, where each of x, y, and z is a positive integer is

(A) 36

(B) 121

(C) $10^{3}-10$

(D) $C_{3}^{10}-C_{2}^{10}$, which denote binomial coefficients

Problem 4:

The hands of a clock are observed simultaneously from 12.45 pm onwards. They will be observed to point in the same direction some time between

(A) 1:03 pm and 1:04 pm

(B) 1:04 pm and 1:05pm

(C) 1:05 pm and 1:06 pm

(D) 1:06 pm and 1:07 pm.

More later,

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