Qq

[prog_tom]

A Point P is chosen at random in the interior of equilateral triangle ABC. What is the probability that ABP has a greater area than each of ACP and BCP?

1/6?

[prog_tom]

Find

Sin^2 10 + ... + Sin^2 90

I know the answer is 5, but how?

Sin^2 = 1 - Cos^2

... so

it would be

9 - Cos^2 10 - Cos^2 20 - ... - Cos^2 90

... the problem is that I want to evaluate it by hand, not calculator.

[Jared]

For question A:

2/3 are the chances that ABP will have more area the ACP or BCP

[prog_tom]

Can you explain how you got that pls?

[Jared]

Consider triangle ABC, with point D. D is created by bisecting the 3 angle in the equilateral triangle.

The triangle is now divided into six sections.

Now consider ABD. If your point P falls into ABD, then any of the other two angles will be larger in area. If point P falls out of ABD, then you have angle ABP which is outside of ABD and hence, has more area.

The odds of P being outside of ABD are 4 in 6 or 2/3rds.

[Jared]

well, i just proved the sin^2 10 + ... + sin^2 90 = 5 without a calculator....

now i will type up my chicken scratch and post it....

[Jared]

sin^2 10 + ... + sin^2 90 = 5

Step 1.
Evaluate the terms that involve special triangles (ie, 30,60,90).
sin^2 30 = 1/4
sin^2 60 = 3/4
sin^2 90 = 1
so these add up to 2.

Step 2.
Evaluate the non special triangle terms.
There are 6 terms left.
Each pair of terms will add to the sin^2 90, which will equal 1.

I prove the identity of one pair only, the process would be the same for the other two.
Here we go....

The argument is that sin^2 90 (which is 1) is equal to sin^2 40 + sin^2 50

sin^2 40 + sin^2 50 'original
sin^2 40 + (sin 50)(sin 50) 'expand the square
sin^2 40 + (sin 50)(cos 40) 'the sinx = cos (90-x)
sin^2 40 + (cos 40)(cos 40) ' the sinx = cos (90-x)
sin^2 40 + cos^2 40 = 1 'sin^2 x + cos^2 x = 1

Do this same process to the other 2 (70 +20, and 80 +10), you'll get 5!

[riis]

Originally posted by Jared
Consider triangle ABC, with point D. D is created by bisecting the 3 angle in the equilateral triangle.

The triangle is now divided into six sections.

Now consider ABD. If your point P falls into ABD, then any of the other two angles will be larger in area. If point P falls out of ABD, then you have angle ABP which is outside of ABD and hence, has more area.

The odds of P being outside of ABD are 4 in 6 or 2/3rds.

Suppose B' is the point on AC which is intersected by the bisector of B, and A' is the point on BC intersected by the bisector of A.
Now, if P falls within ADB' or BDA', then ABP will always be smaller than BCP or ACP (depending on the situation). The area will only be larger if P is on the other side of both bisectors as seen from AB. So, the chance that ABP will be larger is 1/3. (Aren't there three triangles as well. In an equilateral triangle every triangle has a chance of 1/3 that it will be larger than the other two. )

[Jared]

Yes, I am aware of that....


I understood the wording of the original question to be larger than one or the other triangle, not both.

What is the probability that ABP has a greater area than each of ACP and BCP?

2/3 for greater than one or the other, or 1/3 for greater than both....

Perhaps I should be spending my time in a grammar forum, rather than math, eh?

I don't like words much, so much more is accurately expressed in symbols...