... I share one solution ...

Here's my solution to post #67, page 4, Triangle Problem.

∠EDC = 180° - (∠D + 48°) ≊ 81°

where

∠D = acos[(ED² + BD² - BE²)/(2·ED·BD)] ≊ 51° S-S-S equation

BE = sin(84°)/sin(18°) - sin(63°)/sin(39°) ≊ 1.802517077

ED = BE² + BD² - (2·BE·BD·cos18°) ≊ 0.7167358991 S-A-S equation

BD = sin(78°)/sin(18°) - AD ≊ 2.165352128

AD = 1

BEST!

SlideRule

ps: hopefully, NO typos!

The following should be a bit easier than the last one, still I'm not finished. I do not have the 50g at the moment that is busy computing the tournament simulations.

So, if I am not mistaken, I realized that the angle ABC is bisected in E, that CB and AB are equal, that I can apply the Heron's formula, but then it was too late and I started to do rubbish. I am not sure if I can check the problem again until Monday though. Also, the little math handbook that I use to refresh known relationships is not helping, not even the Heron's formula is there.

(04-06-2017 10:43 PM)pier4r Wrote: [ -> ]... The following should be a bit easier than the last one ...

Please clarify the following:

??= AE, from the illustration. What is AE equal to?

BEST!

SlideRule

(04-06-2017 11:24 PM)SlideRule Wrote: [ -> ]Please clarify the following: ??= AE, from the illustration. What is AE equal to?

CE = AE, look at the image, both are labeled with length 'x'.

Area = 7

.......

Spoiler

........

........

The shape of the outer triangle is irrelevant.

At first I had made the sides of the outer triangle = 2x, 3x and 4x. Then I used the law of cosines on each side of outer triangles (sides proportional to 3, 4 and 5) in order to determine the angles. By applying the law of the cosines again on these angles and the sides of the inner triangle opposed to them, these sides could be expressed in terms of x. The Heron's formula was then used to compute the areas of both triangles. Finally, the ratio between these two areas is used to find the answer to the problem.

But, since the shape of the triangle is irrelevant, why not choosing a right-triangle (now I can do it :-), one with angles 90, 60 and 30 degrees and sides 2x, 4x and 2x*sqrt(3)? This makes the solution even more simple: there's no need to use the Heron's formula to calculate the area of the outer triangle; also, the sides of the inner triangle are more easily obtained.

Done by hand, except the evaluation and simplification of the area of the inner triangle using the Heron's formula (HP 50g CAS).

(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]Area = 7

While I appreciate the contribution, I do believe that the problem can be solved without trigonometry.

(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]...The shape of the outer triangle is irrelevant ...

IF the

base and the

height remain

constant, the AREA of the triangle is also

constant, irrespective of an alteration of the "shape" of the original triangle

(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]... the sides of the outer triangle = 2x, 3x and 4x...

Then why not give the

original dimensions entirely in terms of x? The substitution is plausable; is it implied in the

original set of givens?

I approached my attempted solution in a similar manner by asserting the

irrelevancy of the SHAPE for the AREA of a triangle but got caught up on the x=y=z substitution.

BRILL!

SlideRule

(04-07-2017 09:46 AM)pier4r Wrote: [ -> ] (04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]Area = 7

While I appreciate the contribution, I do believe that the problem can be solved without trigonometry.

So do I. My only solace is that at least a calculator is not needed for this one.

(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]Area = 7

.......

Spoiler

........

........

The shape of the outer triangle is irrelevant.

At first I had made the sides of the outer triangle = 2x, 3x and 4x. Then I used the law of cosines on each side of outer triangles (sides proportional to 3, 4 and 5) in order to determine the angles. By applying the law of the cosines again on these angles and the sides of the inner triangle opposed to them, these sides could be expressed in terms of x. The Heron's formula was then used to compute the areas of both triangles. Finally, the ratio between these two areas is used to find the answer to the problem.

But, since the shape of the triangle is irrelevant, why not choosing a right-triangle (now I can do it :-), one with angles 90, 60 and 30 degrees and sides 2x, 4x and 2x*sqrt(3)? This makes the solution even more simple: there's no need to use the Heron's formula to calculate the area of the outer triangle; also, the sides of the inner triangle are more easily obtained.

Done by hand, except the evaluation and simplification of the area of the inner triangle using the Heron's formula (HP 50g CAS).

I also got 7 for a solution but my solution required merely the insertion of two parallel lines

(04-07-2017 12:57 PM)SlideRule Wrote: [ -> ] (04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]...The shape of the outer triangle is irrelevant ...

IF the base and the height remain constant, the AREA of the triangle is also constant, irrespective of an alteration of the "shape" of the original triangle

Exactly! As long as the base and the height are kept constant, the area won't change.

(04-07-2017 12:57 PM)SlideRule Wrote: [ -> ] (04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]... the sides of the outer triangle = 2x, 3x and 4x...

Then why not give the original dimensions entirely in terms of x? The substitution is plausable; is it implied in the original set of givens?

Although apparently misleading, that might have excluded the right-triangle possibility, which is perhaps the most adequate way to solve the problem (at least using trigonometry). An interesting variation of the problem might be the case CF = BF. The solution is also an integer number.

Gerson.

(04-07-2017 02:38 PM)Han Wrote: [ -> ] (04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]Area = 7

.......

Spoiler

........

........

The shape of the outer triangle is irrelevant.

At first I had made the sides of the outer triangle = 2x, 3x and 4x. Then I used the law of cosines on each side of outer triangles (sides proportional to 3, 4 and 5) in order to determine the angles. By applying the law of the cosines again on these angles and the sides of the inner triangle opposed to them, these sides could be expressed in terms of x. The Heron's formula was then used to compute the areas of both triangles. Finally, the ratio between these two areas is used to find the answer to the problem.

But, since the shape of the triangle is irrelevant, why not choosing a right-triangle (now I can do it :-), one with angles 90, 60 and 30 degrees and sides 2x, 4x and 2x*sqrt(3)? This makes the solution even more simple: there's no need to use the Heron's formula to calculate the area of the outer triangle; also, the sides of the inner triangle are more easily obtained.

Done by hand, except the evaluation and simplification of the area of the inner triangle using the Heron's formula (HP 50g CAS).

I also got 7 for a solution but my solution required merely the insertion of two parallel lines

Euclides, the one with a moustache in the photo, was my math teacher in high school. I think I was sleeping when he was explaining the Elements. I paid more attention to his trigonometry classes, though :-)

It would be nice if there was a feature to hide parts of a post until a user explicitly clicks a spoiler button. Anyway draw a line through E and parallel to AB. Let G be the point of intersection of this line with BC and note G is the midpoint of BC. EGC has area 1/4*24=6. Note that EG has length 1.5y. Hence ADE has area 4. BDF has half the height as ADE (a second line through F parallel to AB helps to see this) but twice the base length. Therefore its area is also 4. Lastly EFG (compared vs BDF) has area 3 using a ratio of their bases.

(04-07-2017 04:11 PM)Han Wrote: [ -> ]It would be nice if there was a feature to hide parts of a post until a user explicitly clicks a spoiler button. Anyway draw a line through E and parallel to AB. Let G be the point of intersection of this line with BC and note G is the midpoint of BC. EGC has area 1/4*24=6. Note that EG has length 1.5y. Hence ADE has area 4. BDF has half the height as ADE (a second line through F parallel to AB helps to see this) but twice the base length. Therefore its area is also 4. Lastly EFG (compared vs BDF) has area 3 using a ratio of their bases.

Very nice!

Hidden text (sort of)
(04-06-2017 10:43 PM)pier4r Wrote: [ -> ]The following should be a bit easier than the last one, still I'm not finished. I do not have the 50g at the moment that is busy computing the tournament simulations.

img: http://i.imgur.com/vuvJwSP.png

So, if I am not mistaken, I realized that the angle ABC is bisected in E, that CB and AB are equal, that I can apply the Heron's formula, but then it was too late and I started to do rubbish. I am not sure if I can check the problem again until Monday though. Also, the little math handbook that I use to refresh known relationships is not helping, not even the Heron's formula is there.

Can you explain the deduction process that leads to the conclusion that CB and AB are equal? I suspect that these are not necessarily equal.

(04-07-2017 12:57 PM)SlideRule Wrote: [ -> ] (04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]...The shape of the outer triangle is irrelevant ...

IF the base and the height remain constant, the AREA of the triangle is also constant, irrespective of an alteration of the "shape" of the original triangle

(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote: [ -> ]... the sides of the outer triangle = 2x, 3x and 4x...

Then why not give the original dimensions entirely in terms of x? The substitution is plausable; is it implied in the original set of givens?

I approached my attempted solution in a similar manner by asserting the irrelevancy of the SHAPE for the AREA of a triangle but got caught up on the x=y=z substitution.

BRILL!

SlideRule

I don't think we can assume that x=y=z (which is the implication in assuming the sides are 2x, 3x, 4x).

EDIT (Additional comments):
The \( n \)-dimensional volume of an object in \( \mathbb{R}^n \) (Euclidean space) is invariant under unimodular transformations (i.e. transformations of the form \(\mathbf{x} \to \mathbf{A} \cdot \mathbf{x} \) with \( |\mathbf{A}| =1 \)). So volume (in this case, 2-dimensional "volume" is really area) remains the same; but lengths are not preserved. On the other hand, relative positions are preserved. For example, since F is 1/4 the distance of BC from the point B, it will remain 1/4 the distance from B under any unimodular transformation, and B, F and C will remain collinear. The assumption that the outermost sides are 2x, 3x, and 4x, while in my opinion incorrect, do not affect the final result because (in the end) we are looking at ratios of lengths (i.e. relative positions of points on a line segment).

(04-07-2017 04:11 PM)Han Wrote: [ -> ]It would be nice if there was a feature to hide parts of a post until a user explicitly clicks a spoiler button.

You could use a code box and place the hidden text after 11 or 12 blank lines:

Code:

Spoiler alert:

Scroll down to view the hidden text

.

.

.

.

.

.

.

.

--------hidden text starts here-------

This is your secret message ;-)

--------hidden text ends here-------

However, I am not sure if the code box gets a scroll bar if it contains more than exactly 11 lines under all circumstances.

Dieter

(04-07-2017 05:19 PM)Han Wrote: [ -> ] (04-07-2017 12:57 PM)SlideRule Wrote: [ -> ]IF the base and the height remain constant, the AREA of the triangle is also constant, irrespective of an alteration of the "shape" of the original triangle

Then why not give the original dimensions entirely in terms of x? The substitution is plausable; is it implied in the original set of givens?

I approached my attempted solution in a similar manner by asserting the irrelevancy of the SHAPE for the AREA of a triangle but got caught up on the x=y=z substitution.

AB and AC must not be colinear being the only restriction, I think.

BRILL!

SlideRule

I don't think we can assume that x=y=z (which is the implication in assuming the sides are 2x, 3x, 4x).

Yes, we can! In fact that was my first approach. I've found cos Â = -1/4, cos B = 7/8 and cos Č = 11/16, the sides of the inner triangle being 2sqrt(5/8)x, sqrt(3/2)x and sqrt(47/8)x. Incidentally the area of the inner triangle = 7.

AB and AC must not be collinear. That's the only restriction I am aware of.

(04-07-2017 05:41 PM)Gerson W. Barbosa Wrote: [ -> ]Yes, we can! In fact that was my first approach. I've found cos Â = -1/4, cos B = 7/8 and cos Č = 11/16, the sides of the inner triangle being 2sqrt(5/8)x, sqrt(3/2)x and sqrt(47/8)x. Incidentally the area of the inner triangle = 7.

AB and AC must not be collinear. That's the only restriction I am aware of.

My point was that we cannot assume they are equal based on the problem statement. A solution that involves the lengths of the segments must include some explanation why a change in the lengths of the segments does not alter the final outcome. I am referring to the comment about how the "shape" of the triangle does not matter (I edited a previous comment to add some discussion on how volume is invariant under transformations whereas length is not.)

Additional comments:

Your solution suggests that the inner triangle is isosceles -- this is not true in the general case.

Yes, the assumption x=y=z appears problematic.

IF x=y=z is true, then the original depiction of the inscribed triangle to the circumscribed triangle is curious (see my depiction).

I'm soooooooooo confused!

[attachment=4657]

BEST!

SlideRule