# one series question and one geometry question ^^

• Aug 14th 2007, 11:27 PM
shosho
one series question and one geometry question ^^

1)simplify:

(2/1!)-(3/2!)+(4/3!)-(5/4!)+...+(1998/1997!)-(1999/1998!)

where k! is teh product of all integers from 1 up to k

AND

2) P and Q are the point on the sides AB and BC of a triangle ABC respectively such that BP=3PA and QC=2BQ. K is the midpoint of the segment PQ. Prove that the area of the triangle AKC is equal to 11S/24 where S is teh area of the triangle ABC

thanx everyone
• Aug 15th 2007, 03:20 AM
earboth
Quote:

Originally Posted by shosho

...

2) P and Q are the point on the sides AB and BC of a triangle ABC respectively such that BP=3PA and QC=2BQ. K is the midpoint of the segment PQ. Prove that the area of the triangle AKC is equal to 11S/24 where S is teh area of the triangle ABC

Hello,

1. Draw a sketch (see attachment)

2. The area of the triangle is:

$a_{ABC}=\frac{1}{2} \cdot AB \cdot h_{AB} = \frac{1}{2} \cdot BC \cdot h_{BC}$

3. Now calculate the areas of the triangles APK, PBQ and QCK. Subtract the values of the areas of these triangles from the value of the area of triangle ABC:

i) $a_{PBQ}=\frac{1}{2} \cdot \frac{3}{4} \cdot AB \cdot \frac{1}{3} \cdot h_{AB} = \frac{1}{4} \cdot a_{ABC}$

ii) $a_{APK}=\frac{1}{2} \cdot \frac{1}{4} \cdot AB \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot h_{AB} = \frac{1}{24} \cdot a_{ABC}$

III) $a_{CQK}=\frac{1}{2} \cdot \frac{2}{3} \cdot BC \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot h_{BC} = \frac{1}{4} \cdot a_{ABC}$

3. Thus area of triangle AKC is:
$a_{AKC} = a_{ABC} - \frac{1}{4} \cdot a_{ABC} - \frac{1}{24} \cdot a_{ABC} - \frac{1}{4} \cdot a_{ABC} = \frac{11}{24} \cdot a_{ABC}$
• Aug 15th 2007, 05:21 AM
ThePerfectHacker
Quote:

Originally Posted by shosho

1)simplify:

(2/1!)-(3/2!)+(4/3!)-(5/4!)+...+(1998/1997!)-(1999/1998!)

I get ,
1997/1998!
• Aug 16th 2007, 12:21 AM
shosho
could you explain how you got this please? im reealy confused on this question seems impossible.

however, i went on another site and did some research (wasnt very helpful) but i picked up on how some questions similar to this came up with really simple ansers like 0 or 1...i was thinking the answer may be something like that:confused:
• Aug 16th 2007, 01:22 AM
earboth
Quote:

Originally Posted by shosho
could you explain how you got this please? im reealy confused on this question seems impossible.

however, i went on another site and did some research (wasnt very helpful) but i picked up on how some questions similar to this came up with really simple ansers like 0 or 1...i was thinking the answer may be something like that:confused:

Hello,

your assumption seems to be right. I can't give you an explanation. But I used a computer to calculate the sum. The result after nearly 6 seconds was 1.

I've attached a screen-shot.
• Aug 16th 2007, 05:31 AM
ThePerfectHacker
Quote:

Originally Posted by shosho
could you explain how you got this please? im reealy confused on this question seems impossible.

however, i went on another site and did some research (wasnt very helpful) but i picked up on how some questions similar to this came up with really simple ansers like 0 or 1...i was thinking the answer may be something like that:confused:

Sorry, I made a minor mistake. But Earboths answer is really really close.

Notice that, for $n$ even,
$\left( \frac{2}{1!} - \frac{3}{2!} \right) + ... + \left( \frac{n}{(n-1)!} - \frac{n+1}{n!}\right) = \frac{n!-1}{n!}$

We prove this by induction.
It is trivially true for $n=1$.

Say that,
$\left( \frac{2}{1!} - \frac{3}{2!} \right) + ... + \left( \frac{n}{(n-1)!} - \frac{n+1}{n!}\right) = \frac{n!-1}{n!} = \frac{n!-1}{n!}$
We will show this implies the truth of $n+2$.
Thus,
$\left(\frac{2}{1!} - \frac{3}{2!} + ... + \frac{n}{(n-1)!} - \frac{n+1}{n!}\right) + \frac{n}{(n+1)!}-\frac{n+1}{(n+2)!}$
The terms in paranthesis sum to,
$\frac{n!-1}{n!} + \frac{n}{(n+1)!}-\frac{n+1}{(n+2)!} = \frac{(n+2)!-1}{(n+2)!}$

$1-\frac{1}{1998!}$
I can see why Earboths calculator said 1.
• Aug 18th 2007, 12:13 PM
ticbol
Quote:

Originally Posted by shosho

2) P and Q are the point on the sides AB and BC of a triangle ABC respectively such that BP=3PA and QC=2BQ. K is the midpoint of the segment PQ. Prove that the area of the triangle AKC is equal to 11S/24 where S is teh area of the triangle ABC

I recalled this Geometry problem in a flash while I'm watching a DVD movie. Something in the movie, the old warpic Never So Few, may have triggered the sudden linkage.

Anyway, I got as far as solving then the relationship between the areas of triangles ABK and CBK but I couldn't relate them to the area of triangle ABC. Now, this time, I got it. Which goes to show once again that solving Math problems depends on the mood of your brain. If the key/keys to the solution don't show up today, maybe they will after some time when your brain is relaxed or is relaxing.

The main key here is the area of any triangle is half of the product of a side and the perpendicular height of the triangle based on that side---even if that perpendicular height does not fall on the said side when projected.

Draw the figure on paper.

In triangle PBQ, it is given that PK=Qk, so BK is a median of triangle PBQ. Since a median bisects a triangle, then triangles PBK and QBK are equal in areas, and are each half of the area of triangle PBQ.

------------------------------------------------------------
Relation of the areas of triangles PBQ and ABC:

Using Area = (1/2)(side1)(side2)(sine of included angle),
Area of triangle ABC, S = (1/2)(AB)(BC)sin(angle B) ----(i)

Area of triangle PBQ = (1/2)(BP)(BQ)*sin(angle B)
Area of triangle PBQ = (1/2)[(3/4)(AB)][(1/3)(BC)]sin(angle B)
Area of triangle PBQ = (1/8)(AB)(BC)sin(angle B) -----------------(ii)

Therefore, comparing (i) and (ii), Area of triangle PBQ is (1/4)S. -------***

So, (Area of triangle PBK) = (Area of triangle QBK) = (1/2)(S/4) = S/8 ------***

------------------------------------------------------------------------------
In triangle ABK:
Let h = perpendicular height of triangle ABK on base AB.
Then, h is also the perpendicular height of triangle PBK on base PB.
And, h is also the perpendicular height of triangle PKA on base PA.
So,
Area of triangle PBK = (1/2)(PB)(h) = (1/2)(3PA)(h) ------(iii)
Area of triangle PKA = (1/2)(PA)(h) ----------------------(iv)
So, comparing (iii) and (iv),
Area of triangle PKA = (1/3)(Area of triangle PBK) = (1/3)(S/8) = S/24 ------***

Likewise, In triangle CBK:
Let y = perpendicular height of triangle CBK on base CB.
Then, y is also the perpendicular height of triangle QBK on base QB.
And, y is also the perpendicular height of triangle QKC on base QC.
So,
Area of triangle QBK = (1/2)(QB)(y) -------------------(v)
Area of triangle QKC = (1/2)(QC)(y) = (1/2)(2QB)(y) -----(vi)
So, comparing (v) and (vi),
Area of triangle QKC = (2)(Area of triangle PBK) = (2)(S/8) = S/4 ------***

----------------------------------------------
Now the Area of triangle AKC is the only unknown.

Area of triangle AKC = (Area of triangle ABC) minus (Areas of triangles PBQ, PKA and QKC)
Area of triangle AKC = (S) - (S/4 +S/24 +S/4)
Area of triangle AKC = (S) - [(6S +S +6S)/24]
Area of triangle AKC = (S) -(13S /24)
Area of triangle AKC = 11S /24

Therefore, proven.