Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

5 problems

1.Positive integers m and n satisfy

Find

2.Find the positive integer n such that

3.A trapezoid has side lengths

and

in some order. Find its area

4.A rectangle with integer side lengths has the property that its area minus 5 times its perimeter equals

Find the minimum possible perimeter of this rectangle.

5.For real numbers a, b, and c, the roots of the polynomial

form an
arithmetic progression. Find

Last edited by tony123 (2024-03-14 21:20:40)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,378

Re: 5 problems

Thanks Tony,

More (hopefully) later.

Bob

ps. Still stuck with http://www.mathisfunforum.com/viewtopic.php?pid=432520#p432520

Maybe I need a little hint please.

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,378

Re: 5 problems

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 58

Re: 5 problems

For Q4, is it the area minus (5 times the perimeter),

As in A - 5P = 2023

Or is it (the area minus 5) times the perimeter,

As in (A - 5) * P = 2023

I'm guessing it's the first one??

Last edited by Keep_Relentless (2024-05-08 23:03:09)

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"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 58

Re: 5 problems

Last edited by Keep_Relentless (2024-05-08 23:02:32)

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"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 58

Re: 5 problems

Oh I forgot the sides are integers.

Last edited by Keep_Relentless (2024-05-08 23:32:24)

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"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 58

Re: 5 problems

I think for Q3 you use Brahmagupta's formula

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"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 58

Re: 5 problems

Regarding Q5, help is needed on finding roots of quintic equations and what's needed to see the answer, thanks

Last edited by Keep_Relentless (2024-05-09 00:33:29)

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"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 58

Re: 5 problems

Can anyone show how to do Q5??

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"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,378

Re: 5 problems

I called the roots a, a+d, a+2d, a+3d and a+4d and expanded that quintic. There were things about the resulting expression that suggested a + b  + c = 9, but I'm not confident that's the correct answer so this one is still on my 'to-do list'.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 184

Re: 5 problems

5.For real numbers a, b, and c, the roots of the polynomial

form an
arithmetic progression. Find

This exercise has a solution if the right side of the equation is 32 instead of 320.

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Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 184

Re: 5 problems

Well, I also found the solution in case the right side is 320.
a+b+c=49

Kerim

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Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 58

Re: 5 problems

Interesting, any chance you can explain how you worked it out? We all had a go and didn't get there.

K_R

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"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 184

Re: 5 problems

Can anyone show how to do Q5??

(x-A)*(x-B)*(x-C)*(x-D)*(x-E) = 0

x^5 - (x^4)*(A + B + C + D + E) + (x^3)*(A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E) - (x^2)*(A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E) + x*(A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E] - A*B*C*D*E = 0
or
x^5 - (x^4)*(A + B + C + D + E) + (x^3)*(A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E) - (x^2)*(A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E) + x*(A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E] = A*B*C*D*E

We have:
x^5 - 10x^4 + ax^3 + bx^2 + cx = 320

By comparing, we get
A + B + C + D + E = 10
A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E = a
A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E = b
A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E = c
A*B*C*D*E = 320

A + B + C + D + E = 10

In general, Sum = [2*m + (n-1)*r]*n/2
where
m is the first term
r is the common difference
n the number of terms

[2*A + (5-1)*r]*5/2 = 10
A + 2*r = 2
r = (2 - A)/2

For A = -4 , r = 3
Therefore
A = -4
B = -1
C =  2
D = 5
E = 8

Their sum is 10
And their product is 320

a = A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E = -5
b = A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E = 190
c = A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E = -136

Therefore:
a + b + c = 49

Kerim

I did a mistake (sorry, a mistyping, not about the asnswer); will someone correct it?

Last edited by KerimF (2024-05-14 01:41:52)

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Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.

Europe2048

Member

Registered: 2024-01-03

Posts: 35

Re: 5 problems

KerimF, use the hide tag if you don't want to straight-up reveal the answer.