IIT JEE , AIEEE Forum - Mathematics , Algebra

magiclko

A triangle has its sides 4, 4 and 5. A point P is taken inside it such that its length of perpendiculars to the side are p1, p2, p3, then find the maximum value of p1.p2.p3

magiclko

no answer..

....pls experts i want it by tomorrow....

puneet

hiiiiiiiiii magicklo ..

Here i am to answer ur question ..

I might not be able to answer ur question perfectly as i cannot draw diagrams but i will tell u the steps, you should simultaneously draw and see for urself .. things will become clearer for sure ...

So here we start ..

First of all I will like to point that since all sides are know so the triangle is completely known to us ..

So, all the angles are fixed and all sides are already given to us ..

okies .. now let us get to actual point ..

Now, see let the 5 side be called the BC and let the perpendicular frm the vertex A meet the BC in point D ..

we call the point D as the origin ..

Now the coordinates of the point P be (x,y)

Now clearly the perpendicular to side BC is of length y

So, p1 = y ..

Now the perpendicular to sides AB and AC are of length (d+x)cosecB and (d-x)cosecB respectively ...

where d is 5(4sinB-y)/8.sinB

Now u may be wondering from where did this d came from ..

Well at the point P draw a staright line parralel to the side BC.

This line meets side AB in H and AD in point K ... so now d is HK ..

How to find HK ?? well just use the concept of similar trianlges we know the lenght of AD = 4sinB and also of KD = y .. also BD = 5/2 ...

So I guess u can find HK easily using similar triangles ... okay

I hope u are getting wat i am saying ... draw a diagram if u have not drawn it ... things will become more clear then ...

Now once we have calculated HK .. p2 and p3 can be calculated easily using simple trignometry ..

and also using the fact that

AHK is

B only (since HK is parralel to BC)

So finally we have the value of all perpendiculars ..

so , p1.p2.p3 = (d+x)cosecB.(d-x)cosecB.y

= (d² - x²)cosec²B.y

Now we know that d is a function of y and has nothing to do with x

Now substractin x² from d² will only reduce the value .. and so for maximum value x must be zero .. i hope this point is clear ..

so, x is zero .. thus now p1.p2.p3 = d²cosec²B.y

=(5(4sinB-y)/8.sinB)²cosec²B.y

= k.y.(y²- 8.sinB.y + 16.sin²B)

where k is 25/64.cosec⁴B ( a constant )

To maximize p1.p2.p3 we need to maximize y.(y²- 8.sinB.y + 16.sin²B)

Let us differntiate g = y.(y²- 8.sinB.y + 16.sin²B) w.r.t y and put it to zero to find the maxima ..

dg/dy = 3y²- 16.sinB.y + 16.sin²B = 0

so, y = 4.sinB , 4/3.sinB

Now , figure out that ... 4/3.sinB will give a non zero value for g = 256/27.sin³B..

okies ... so now value of p1.p2.p3 is 25/64.cosec⁴B.256/27.sin³B

= 100/27.cosecB

I hope this is clear ... bare with me if I had made some calculation mistake .. but the idea is simple ..

cheers

magiclko

thanx for answering sir...... but one thing sir here, u hav taken the perpendiculars frm the vertex, while p1, p2 and p3 are the length of perpendiculars frm point P to the sides...... i dnt knw exactly the answer but m sure its not.... its actually a question of progressions, n so will be done thru AM>GM>HM...... plsss reply again soon.....

puneet

Hii magicklo ..

pls note the answer once again .. I had earlier saved a incomplete answer as I had to go in between ... sorry for the inconvinience caused ..

Get back to me if u do not understand anything

cheers

mohit1

Respected sir, I am getting confused on what you have mentioned on your line no. (9) and (12) . The problem is that- have you taken the perpendiculars being dropped from the opposite vertices or the pt P?

mohit1

no reply??

mohit1

Puneet sir pl explain me also........

magiclko

sir the perpendiculars are frm point P not frm vertices..... so answer will be different..... moreover it was an objective question, it cant be so lengthy!!!

puneet

hiiiiii folks ...

Well the perpendiculars are from P only .. read the solution carefully ..

The question is not lenghty .. it is simple ..

Just that I used this trick to explain .. so it became lenghty .. u shud appreciate that if I cud draw a diagram .. then the solution wud have become relatively small .. I hope u agree on to this ..

Other way cud be to use A.M. > G.M. property ... however u will have to do same things as what I have done ... some small modifications here and there ..

I hope it is clear now ..

cheers

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