Which one is greater e raised to the power π ( e^π) or π raised to the power e ( π^e)?

(For understanding the number e, you may wish to click on this link: e for eternity. And to understand the number π, you may wish to click on the link: True life of π.)

Today's Question: Which one is greater: e raised to the power π (e^π ) or π raised to the power e (π^e)? 

This a classical question.The answer can be found immediately by using Calulator or by taking logarithm of both the numbers and then comparing. 

However, let us try to guess the answer without calculating and without using Logarithm.

                Approach (1) 

In e^π , the base is smaller (e= 2.718), but the exponent is larger (π = 3.14). And in π^e , the base is larger, but the exponent is smaller (e).

But note that both the numbers are in the vicinity/ proximity of 3.

So which effect wins: the larger base or the larger exponent?

Let us try to understand with the example of 2³ and 3². Both these bases/ exponents are 3 or less than 3. Here 3² is greater than 2³. In this case the number with greater base is greater.

Now  take example 3⁴ and 4³, or 4⁵ and 5⁴,  or 5⁶ and 6⁵ so on and so forth. In all these cases numbers with higher exponents are greater.

The key idea here is: For numbers  greater than 3, increasing the exponent has a  stronger effect than increasing the base by a similar amount.

Both changes increase the value, but the gain from increasing the exponent is  larger than the gain from increasing the base. 

From this approach , it is quite safe to conclude that e^π is the greater number as the exponent π is greater than 3. So e^π ends up ahead.

                Approach (2)

Imagine money growing by compound interest: The base is like the interest rate and the exponent is like the number of years you let it compound.

A slightly lower interest rate over more years can outperform a slightly higher interest rate over fewer years. Here, the "extra compounding time" provided by the exponent π more than compensates for the slightly smaller base e.

So, in plain language: Although π is bigger than e, the advantage of having the larger exponent (π instead of e) outweighs the advantage of having the larger base.

Therefore, e^π  is larger than π^e .

Concluding Part:   

A fascinating historical note is that π and e arise from completely different origins.

π comes from geometry: the ratio of a circle's circumference to its diameter. Whereas e comes from growth and continuous compounding, calculus, and logarithms. But they both gel so well in different situations.

We will discuss more about the gelling of e and π soon in a different context.         

Ptolemy's Theorem and Golden Ratio.

The Golden Ratio has been a recurring star on this blog ever since we first explored its mysterious beauty years ago. If you missed that journey, be sure to click here for the original article.

Its magic did not end there. We went on to solve a whole collection of fascinating problems based on the Golden Ratio, each revealing yet another surprising facet of this extraordinary number. You can revisit those collection by clicking here.

More recently, we explored the elegance of Ptolemy’s Theorem and walked through its proof in detail. Now comes the exciting part!

Today, we shall bring these two mathematical gems together. With a single, elegant application of Ptolemy’s Theorem, we will derive one of the most celebrated results in geometry: the ratio of the length of a diagonal of a regular pentagon to the length of its side. As you will soon discover, that ratio is none other than the magnificent Golden Ratio which is equal to (√5 +1)/2 = 1.618.

Please note that any regular polygon can be inscribed in a circle. So any four vertices of a regular pentagon taken in order will be forming a Cyclic Quadrilateral. And you know the famous Ptolemy's Theorem for cyclic quadrilaterals.

Now, let us draw the diagram by hand. (Computer-generated figures may be perfect, but handwritten sketches possess a soul of their own. Besides, I would much rather leave this world having left not only erasable footprints on the dust of Indoree streets, but fingerprints also on the pages I cherished!) So here is the Diagram and the derivation.

The diagram is self explanatory.

By applying this theorem, one arrives at a relationship of such striking elegance and simplicity that, once encountered, it lingers in the mind forever; you could scarcely forget it, even if you tried.

Though seldom encountered beyond the confines of mathematical circles; this elegant property is a hidden treasure: one that rewards serious puzzle lovers and proves immensely useful for students preparing for competitive examinations such as XAT, CAT, and their counterparts.

Finally, let us apply this knowledge of Golden Ratio to solve this quiz in circulation in quiz groups:

The problem is stated as below: Find value of x without use of Trigonometry/ Trigonometric Tables.

While this problem can be solved by various geometrical methods, we will solve it with the help of above property of regular Pentagon that  the ratio of the Diagonal to the side is equal to Golden Ratio which is equal to (√5 +1)/2 = 1.618.

Angle 36° is naturally found in Regular Pentagons. Let the triangle AFE given in the problem be a part of a regular pentagon of side 2 as shown in the attached image.

In the diagram, AE=2, so BE will be equal to 2x (√5 +1)/2 = √5+1. Hence FE or x will be equal to (√5+1)/2.

Thus we can find the value of x instantly after seeing the problem!

Similarly this problem can be solved in the similar manner by constructing 36°-72°-72°  or 36°-36°-108° triangle which can be found in Regular Pentagons. (Click here to know more about 36°-72°-72° triangle. And click here to know more about 36°-36°-108° triangle.).

Ptolemy's Theorem and its Proof.

Before we discuss about Ptolemy's Theorem,

let us first try to know who Ptolemy was.

Claudius Ptolemy was an influential mathematician, astronomer, and geographer who lived in Alexandria during the second century CE. He is best known for Ptolemy’s Theorem in geometry and his geocentric model of the universe, which shaped scientific thought and astronomical studies for over a thousand years.

        Ptolemy's Theorem

In any Cyclic Quadrilateral (a quadrilateral whose four vertices lie on the same circle), the product of the lengths of the two diagonals is equal to the sum of the products of the lengths of the two pairs of opposite sides.

We all encountered this theorem somewhere in our school Geometry syllabus. Whether our teachers skipped over it, or Students quietly skipped over it, is another story. Yet it was always there, waiting patiently in the textbook. A few curious minds discovered its elegance back then, while many others are only now getting the chance to appreciate the amazing beauty hidden within it.

Now let us explore a beautiful proof of Ptolemy’s Theorem using the elegant concept of similar triangles. The argument is remarkably simple, easy to follow, and reveals the theorem’s hidden harmony with surprising clarity.

Interestingly, Ptolemy’s Theorem can also be proved using the celebrated Pythagoras’ Theorem; another timeless result that is both familiar and strikingly easy to understand. It is fascinating to see how one beautiful theorem can naturally lead to the proof of another.👇

Ptolemy’s Theorem is far more than a classical result in Geometry. It has appeared, directly or indirectly, in numerous problems from competitive examinations such as CAT and XAT, where a quick application of the theorem often leads to an elegant and time-saving solution.

There exists a myriad of geometrical problems that yield almost effortlessly to the elegant application of such theorems, often revealing their solutions in the blink of an eye.

In the next blog post, we shall explore a particularly elegant and useful result that emerges naturally from the application of Claudius Ptolemy's theorem, further illustrating its enduring beauty and the surreal Joy of Mathematics.

36°–72°–72° Triangle

Among all special triangles in geometry, the 36°–72°–72° triangle occupies a unique place. At first glance it appears to be just another isosceles triangle, but hidden within it lies one of mathematics' most celebrated numbers: The Golden Ratio.The ratio of an equal side to the base of a 36°–72°–72° triangle is exactly the Golden Ratio (√5+1)/2 =1.618

The 36°–72°–72° triangle naturally appears inside a regular pentagon and a pentagram. Every time diagonals are drawn in a regular pentagon, these elegant triangles emerge, carrying with them the Golden Ratio in a surprisingly effortless way.

This intimate connection explains why the Golden Ratio and pentagonal geometry are inseparable in classical mathematics.

Splitting the triangle along its axis of symmetry creates two right triangles with angles 18°–72°–90°. It frequently appears in geometric puzzles, olympiad problems, and CAT-style aptitude questions where trigonometry can often be avoided through pure geometric reasoning.

The 36°–72°–72° triangle is a perfect example of hidden mathematical elegance and Mathematicians/Puzzle Lovers just love it.

A simple arrangement of three angles leads to irrational numbers, beautiful algebraic identities, and deep links with regular polygons. It serves as a bridge between elementary geometry and the rich world of the Golden Ratio.

Once you recognize this triangle in a figure, many seemingly difficult problems become surprisingly simple. Indeed, it is one of those rare geometric gems that rewards observation more than computation.

All the salient features of 36°–72°–72° triangle are reproduced below in two posters  which can be circulated to students and Maths Lovers alike.

The Elegant 36°–36°–108° Triangle

Geometry is full of hidden treasures, and the 36°–36°–108° triangle is undoubtedly one of them. Although it is simply an isosceles triangle with two equal angles of 36°, it possesses remarkable properties that connect it directly to the Golden Ratio, the regular pentagon, and the beautiful geometry of the pentagram. It is a Natural and Permanent Resident of the Regular Pentagon!

Suppose the two equal sides each have length 1, and let the base have length b.

In a 36°–36°–108° triangle, the ratio of the base to either equal side is exactly the Golden Ratio.

If the equal sides are 1 unit long, the base measures approximately  (√5+1)/2 or 1.618 units.

Take any regular pentagon and draw one of its diagonals. The triangle formed by that diagonal and the two adjacent sides is a 36°–36°–108° triangle.

This explains why the Golden Ratio appears so naturally in pentagonal geometry. In fact, the diagonal of a regular pentagon is exactly (√5+1)/2 or Golden Ratio times as long as one of its sides. 

If we draw the altitude from the 108° vertex to the base.

This divides the triangle into two congruent right triangles, each having angles

18°, 36°, and 90°.These right triangles are themselves rich in elegant identities and often appear in olympiad problems and geometric constructions.

The 36°–36°–108° triangle is a wonderful reminder that extraordinary mathematics can arise from an ordinary looking figure. Its intimate relationship with the Golden Ratio, regular pentagons, and classical geometry makes it a favourite among puzzle enthusiasts and mathematicians alike.

Once you learn to recognize this triangle, you'll begin spotting it in pentagrams, geometric proofs, and competition problems.  And you'll discover that it often unlocks elegant solutions with very little computation.

The various features of 36°–36°–108° Triangle are presented in a poster, which can be sent to students or maths lovers alike.

Sum of Several Consecutive Natural Numbers...

Question: Which numbers can be or can not be expressed as a sum of two or more consecutive natural numbers. Let us divide the solution into a few small parts for quick and better understanding.

                 Part 1

First of all let us see which numbers can not be expressed as a sum of several consecutive natural numbers.

The answer lies in this fact: If we take two consecutive natural numbers; one will be odd and the other will be even, in either order.

But if we consider the numbers which are powers of 2 like 2, 4, 8,16, 32, 64 etc; they or their factors don't have any odd factor (other than 1) so they can not be expressed as a sum of two or more consecutive natural numbers. 

So the necessary condition for a number to be expressed as a sum of several consecutive natural numbers is that it should have at least one odd factor, other than 1.

                 Part 2

Now let us come to odd numbers. 

Every odd number n can be written as (2k + 1) which can further be written as k + (k+1). The numbers k and k+1 are two consecutive numbers.

So all odd numbers can definitely be expressed as a sum of two consecutive natural numbers. Thus all odd numbers are sorted.

                 Part 3

Now we are left with only the even numbers which have at least one odd factor other than 1. Or in other words, now let's find the even numbers that aren’t powers of two.

Clearly, such even numbers can only be the sum of a series of more than two consecutive numbers.

They must have an even number of odd numbers in the series of consecutive numbers, but any number of even numbers.

These even numbers are all of the form 2n.

If n (the half) is an odd number, n can be written as a pair of two consecutive numbers.

Then 2n can also be written as a sequence of four consecutive natural numbers, instead of two numbers.

To demonstrate this, let us take any even number that’s not the power of two and whose half is odd: Say 50.

It's half=25 which can be written as a sum of 12+13.

And 50 can be written as equal to 12+12+13+13 which can be rearranged as 11+12+13+14. We have balanced it by adding and subtracting 1 from two numbers at the end to get the desired result.

The same method of adding 1 and subtracting 1 several times can be applied to any such even number and some series of Consecutive Natural numbers can be easily found.

                Part 4

In fact, we can generate a generalised formula to know the number of ways a number can be written as a sum of two or more consecutive natural numbers. Interestingly, we can also find ‘the ways’ too with a bit of mathematical jugglery.

I have already given the formula way back in 2013 to my beloved students who were also in pursuit of ‘Joy of Mathematics’ apart from being aspirants for CAT-IIM.

It will require revisiting Number Theory and Factor theory. You may wish to click here to revisit my blog post ( Short Cuts for Numbers) on Number theory to understand the formula. The formula is given in the hand written note at the serial number 17. 

We can discuss its full derivation in a separate blog post once you become familiar with the first sixteen formulae as given in the above mentioned blog post.

(You may also wish to solve a few questions on numbers from the Blog Post: Investigations of a Number 420.)

If sufficient friends gather, I can pick up chalk once again for a day or two in a week at any place in Indore.

Let the conventional Joy of Mathematics prevail forever…

A Square in an Egyptian Triangle.

 A question has been received in the image form as below:

While it can be solved by many methods; perhaps it will be better to solve it considering that this triangle is  similar to the Egyptian Triangle ( 3-4-5 Right Triangle).

There is a property of 3-4-5 right triangles that the inner sides of the square inscribed in it divide the sides of the triangle in the ratio of 3:4.

As the hypotenuse of the problem figure is divided in the ratio of 3:4, clearly this is a scaled up  triangle similar to the 3-4-5 right triangle. 

The Hypotenuse is 84/5  times of the original hypotenuse of the Egyptian  triangle. So now we can write the base equal to 4*84/5= 336/5 and perpendicular equal to 252/5.

The upper portion of the perpedicular will be 3/7 of 252/5 and the lower portion will be 4/7 of 252/5.

The left portion of the base will be 3/7 of 336/5 and the right portion will be 4/7 of 336/5.

The whole problem is now reduced to simple calculation as shown in the hand written note produced below:

The whole idea is why to reinvent the wheel! It is existing beautifully for us. We have to make vehicles only to surge ahead.

When we know about the Egyptian Triangle, it is better to apply its properties to such problems.

You may wish to click here to go through the blog post on Egyptian Triangle once again and make your own notes or formulae for such problems.

Angle between the Hands

Let us make a formula to know the exact angle between the minute hand and hour hand of a Clock. Let the angle between the hands be A, no. of hours be H and no. minutes be M.

So we will establish a relationship between A,H and M.

First of all let us determine angular speeds of minute hand and hour hand.

Minute hand travels 360 degrees in 60 minutes. Hence its speed is 360/60 or 6° per minute.

The hour hand being much slower, takes 12 hours or 720 minutes to cover 360 degrees.

Hence  hour hand’s speed = 1/2° per minute or 30° per hour.

Imagine a clock initially showing 0° angle, which occurs exactly at 12 O’ Clock. And also imagine a vertical line between 12 O’ Clock and 6 O’ Clock.

Suppose, the hour hand travels for H complete hours and M Minutes from the initial position, and M hand travels for M minutes from the initial position.

Angular distance covered by Minute Hand = speed x time =6 M

Angular distance covered by Hour Hand = H hours + M minutes= 30H+ M/2.

Now the angle between the hour hand and minute hand will be the difference of the two angular distances 

Please note that for an angle A, sometimes minutes hand will be ahead and sometimes hour hand will be ahead.

Hence the angle A will be equal to = 6M- ( 30H +M/2)= 11M/2-30H  ( This will happen when Minute hand will be ahead)

A= 11M/2-30H

And when Hour hand will be ahead the angle A will be equal to 30H+ M/2 -6M= 30 H-11M/2

A= 30 H-11M/2

We can now write the result in the modulus form, combined for both the cases:

A = |11M/2-30H|

In the future you can just put the values in the above formula and find the unknown quantity.

Even if you don't understand the derivation of the formula, you can still use it smartly.

For example look at the problem given in the figure:

Here hour hand is between 9 and 10. Hence H= 9.

Angle A= 180°.

As the  hour hand is ahead, we will use the second formula.

A= 30 H-11M/2

Putting the values in the above equation, we can find the value of M.

180= 30X9-11M/2

M= 180/11.

  M = 16 and 4/11 minutes.

Normally such  answers are conventionally given in the mixed fraction form only. However the same can be converted into seconds also.

So the final answer will be 16 minutes, 21 and 2/11 seconds.

Or the same may be written as 16 minutes and 21.8181 seconds or 16 minutes and 21.82 seconds. You can say approximately 16 minutes and 22 seconds also.