TODAY'S PROBLEM:AN APPLICATION OF MENALAUS’ THEOREM

Given: AB is the diameter of a circle with center O. C be any point on the circle. OC. is joined. Let Q be the midpoint of OC. AQ produced meet the circle at E. CD be perpendicular to diameter AB. ED and CB are joined.

R.T.P. : CM = MB

Construction: AC and BD joined.

Proof: In triangle BOC, AQF is the transversal. Applying Menalaus’ Theorem we have CF/FB = 1/2.
Now ΔABC is similar to ΔBDH and ΔACF is similar to ΔDGH (right angles equal and angle subtended by the same segment equal in both cases).
Hence HG/GB = CF/FB. But CF/FB =1/2. Hence HG/GB = 1/2.
Now applying Menalaus’ Theorem in ΔBCH with transversal DGM we have CM/MB=1.

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TODAY'S PROBLEM

Let X = {1, 2, 3, … , 10}. Find the number of pairs {A, B} such that A ⊆ X, B ⊆ X, A ≠ B and A∩B = {5, 7, 8}.

Solution:

First we put 5, 7, 8 in each of A and B.

We are left out with 7 elements of X.

For each of these 7 elements there are three choices:

a) it goes to A

b) it goes to B

c) it goes to neither A nor B

Hence there are total \(3^7\) = 2187 choices. From these 2187 cases we delete that one case where all of the seven elements goes to neither A nor B as A≠ B thus giving 2187 -1 = 2186 cases.

Since A and B is unordered (that is A= {5, 7, 8, 1, 2} , B = {5, 7, 8, 4} is the same as B= {5, 7, 8, 1, 2} , A = {5, 7, 8, 4} ) we take half of these 2186 cases that is 1093 cases.

Hence there are 1093 such pairs.

1992 Inmo

INMO 91

TODAY'S PROBLEM

Let  . Prove that .

Discussion:

We consider the function . The first derivative of this function is  In the interval  the numerator is always negative as x is less than tan x.

Hence f(x) is a monotonically decreasing function in the given interval. Hence f(x) attains least value at  which equals 

Therefore  in the given interval.