Test of Mathematics Solution Subjective 110 - Ratio of Diagonals of Cyclic Quadrilateral

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 110 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem:

:
Let ABCD be a cyclic quadrilateral with lengths of sides AB = p , BC = q , CD = r, and DA = s . Show that $ \frac{AC}{BD} = \frac{ps+qr}{pq+rs} $

Solution:

Since ABCD is cyclic $ \Delta ABE $ is similar to $ \Delta CDE $ (since $ \angle ABE = \angle DCE $ as they are subtended by the same arc AD, and $ \angle AEB = \angle CED $ as vertically opposite angles are equal)

Hence their corresponding sides are proportional.

$ \frac{p}{r} = \frac{BE}{CE} \Rightarrow BE = CE \times \frac{p}{r} $
$ \frac{p}{r} = \frac{AE}{DE} \Rightarrow AE = DE \times \frac{p}{r} $

Similarly $ \Delta AED $ is similar to $ \Delta BEC $

$ \frac{s}{q} = \frac{DE}{CE} \Rightarrow DE = CE \times \frac{s}{q} $
$ \frac{q}{s} = \frac{CE}{DE} \Rightarrow CE = DE \times \frac{q}{s} $

Hence
$ BE + DE = BD = CE \times { \frac{p}{r} + \frac{s}{q} } = CE \times \frac{pq+rs}{qr} $
$ AE + CE = AC = DE \times { \frac{p}{r} + \frac{q}{s} } = DE \times \frac{ps+qr}{sr} $

Hence $ \displaystyle { \frac{AC}{BD} = \frac {ps + qr}{pq + rs} \times \frac {qr}{sr} \times \frac {DE}{CE} }$

Finally we note that since $ \Delta AED $ is similar to $ \Delta BEC $. $ \displaystyle {\frac{q}{s} = \frac {CE}{DE} \Rightarrow \frac{q}{s} \times \frac {DE}{CE} = 1 } $

This proves that

$ \displaystyle { \frac{AC}{BD} = \frac {ps + qr}{pq + rs}}$

Test of Mathematics Solution Subjective 37 - The prime 13

Test of Mathematics at the 10+2 Level

Test of Mathematics Solution Subjective 37 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

Supposed p is a prime Number such that (p-1)/4 and (p+1)/2 are also primes. Show that p=13.

Discussion:

p is not 2 or 3 (otherwise (p-1)/4 would not be an integer).Hence p must be an odd prime. Also p-1 is divisible by 4

p = 4t + 1 (say)

(p-1)/4 = t

(p+1)/2 = 2t + 1

Hence t, 2t+1, 4t+1 are all primes.

If t = 3 then these numbers are 3, 7 and 13.

If t is not 3 then t must produce 1 or 2 remainders when divided by 3 (t is a prime, hence cannot be divisible by 3).

$ \displaystyle {t \equiv 1 \text{mod} 3 \Rightarrow 2t +1 \equiv 3 \equiv 0 \text {mod} 3} $

But 2t +1 is a prime. So it is impossible that 3 divides 2t +1. Hence t cannot be 1 mod 3.

Similarly $ \displaystyle {t \equiv 2 \text{mod} 3 \Rightarrow 4t +1 \equiv 9 \equiv 0 \text {mod} 3} $

But 4t +1 is a prime. So it is impossible that 3 divides 4t +1. Hence t cannot be 1 mod 3.

Therefore we can have no other t such that the given condition is satisfied. Hence t must be 3 and the prime must be 13.

Clocky Rotato Arithmetic

Do you know that CLOCKS add numbers in a different way than we do? Do you know that ROTATIONS can also behave as numbers and they have their own arithmetic? Well, this post is about how clock adds numbers and rotations behave like numbers. Let's learn about clock rotation today

Consider the clock on earth.

clock

So, there are 12 numbers {1,2, ..., 12 } are written on the clock. But let's see how clocks add them.

What is 3+ 10 ?

Well, to the clock it is nothing else than 1. Why?

Say, it is 3 am and the clock shows 3 on the clock. Now you add 10 hours to 3 am. You get a 13th hour of the day. But to the clock, it is 1 pm.

So, 3 + 10 = 1.

If you take any other addition, say 9 + 21 = 6 to the clock ( 9 am + 21 hours = 6 pm ).

Now, you can write any other Clocky addition. But you will essentially see that the main idea is :

The clock counts 12 = 0.

Isn't it easy? 0 comes as an integer just before 1, but on the clock, it is 12 written. So 12 must be equal to 0. Yes, it is that easy.

Cayley's Table

This is a handsome and sober way to write the arithmetic of a set. It is useful if the set is finite like the numbers of the CLOCK Arithmetic.

Let me show you by an example.

Consider the planet Cheenta. A day on Cheenta consists of 6 earth hours.

So, how will the clock on Cheenta look like?

Cheenta Planet clock

Let's us construct the Cayley Table for Cheenta's Clocky Arithmetic. Check it really works as you wish. Here for Cheenta Clock, 3 = 0.

Clock rotation explanation

Exercise: Draw the Cayley Table for the Earth (24 hours a day) and Jupiter (10 hours a day).

Nice, let's move on to the Rotato part. I mean the arithmetic of Rotation part.

Let's go through the following image.

clock rotation

Well, let's measure the symmetry of the figure. But how?

Well, which is more symmetric : The Triskelion or the Square (Imagine).

clock rotation

Well, Square seems more right? But what is the thing that is catching our eyes?

It is the set of all the symmetric positions, that capture the overall symmetry of a figure.

For the Triskelion, observe that there are three symmetric operations that are possible but that doesn't alter the picture:

For the Square, the symmetries are:

For, a square there are symmetries, hence the eyes feel that too.

So, what about the arithmetic of these? Let's consider the Triskelion.

Just like 1 interact (+) 3 to give 4.

We say \(r_1\) interacts with \(r_2\) if \(r_1\) acts on the figure after \(r_2\) i.e ( 240 + 120 = 360 degrees rotation = \(r_3\) ).

Hence, this is the arithmetic of the rotations. To give a sober look to this arithmetic, we draw a Cayley Table for this arithmetic.

Treskelion

Well, check it out.

Exercise: Can you see any similarity of this table with that of anything before?

Challenge Problem: Can you draw the Cayley Table for the Square?

You may explore this link:- https://cheenta.com/tag/level-2/

And this video:- https://www.youtube.com/watch?v=UaGsKzR_KVw

Don't stop investigating.

All the best.

Hope, you enjoyed. 🙂

Passion for Mathematics.

Consecutive composites | TOMATO Objective 151

This is an objective problem 151 from TOMATO based on Consecutive composites, useful for Indian Statistical Institute Entrance Exam.

Let $n = 51! + 1$. Then the number of primes among $n+1, n+2, ... , n+50$ is

(A) $0$;

(B) $1$;

(C) $2$;

(D) more than $2$;

Discussion:

$51!$ is divisible by $2, 3,... 51$.

Hence $51! +2$ is divisible by $2, ... , 51! + k$ is divisible by k if $k \le 51 $

Therefore all of these numbers are composite (none of them are primes).

Answer is (A) 

Note: The above problem can be further extended to say that for any natural number of n, we can have n consecutive composite numbers.