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(08-10-2015 12:07 AM)CR Haeger Wrote: [ -> ]... If this is correct ...
... I expect the 50g or Prime could make quick work of this.

This is correct.
Prime users aren't engaged in this little things. They go to sleep and don't know, whether their calculator will work tomorrow.
(08-10-2015 02:12 PM)Hlib Wrote: [ -> ]
(08-10-2015 12:07 AM)CR Haeger Wrote: [ -> ]... If this is correct ...
... I expect the 50g or Prime could make quick work of this.
Prime users aren't engaged in this little things. They go to sleep and don't know, whether their calculator will work tomorrow.

I'm a Prime user too... It seemed to work (quickly) today. I am a novice on the 50G.

Using CAS: solve([(2*x-3*y) = (6-7*a^2),(-3*x+5*y) = (7+2*a),(7*x+11*y) = (4-6*a)],[x,y,a])

Exact x,y,a: {[(-52*√21106-7381)/8092,(-4*√21106+7213)/8092,(2*√21106+23)/238],[(52*√21106-7381)/8092,(4*√21106+7213)/8092,(-2*√21106+23)/238]}

Approx x,y,a: {[−1.84571307168,0.819560532948,1.31747093989],[0.021442186857,0.963187860527,−1.12419362897]}
(08-10-2015 02:12 PM)Hlib Wrote: [ -> ]Prime users aren't engaged in this little things. They go to sleep and don't know, whether their calculator will work tomorrow.
My Prime has been working every day I needed it, so I don't fear tomorrow For the first problem (sum of factors of 224551579230):
Code:
CAS.ifactors(224551579230)▶L1 ΣLIST(MAKELIST(L1(I)*L1(I+1),I,1,SIZE(L1),2))
returns 301

And for the second problem (2015 .19), in CAS mode:
Code:
solve({2x-3y=6-7a,-3x+5y=7+2a,7x+11y=4-6a},[a,x,y])
returns a=235/128  and with a^2 in the first equation :
Code:
 solve({2x-3y=6-7a^2,-3x+5y=7+2a,7x+11y=4-6a},[a,x,y])
returns for a: (-2*√21106+23)/238 & (2*√21106+23)/238  The following problem, from the 2013 Houston calculator exam, seems conceptually simple to solve, but the manipulations necessary to solve it from start to finish on the 50g are not so obvious:

14. There are 3 points of intersection of pairs of the lines 2x+3y=4, 7x-2y=3 and
x+y=15. Give the area of the triangle determined by these three points.

It seems clear that the "shoelace formula" of Meister and Gauss* is needed, and a little program to apply that would be nice.

Peter

* <https://en.wikipedia.org/wiki/Shoelace_formula>
(08-12-2015 04:22 PM)Peter Murphy Wrote: [ -> ]It seems clear that the "shoelace formula" of Meister and Gauss* is needed, and a little program to apply that would be nice.

Cf. 50G Question, has anyone programmed gauss's...

Cheers
Thomas
(08-12-2015 04:22 PM)Peter Murphy Wrote: [ -> ]The following problem, from the 2013 Houston calculator exam, seems conceptually simple to solve, but the manipulations necessary to solve it from start to finish on the 50g are not so obvious:

14. There are 3 points of intersection of pairs of the lines 2x+3y=4, 7x-2y=3 and
x+y=15. Give the area of the triangle determined by these three points.

It seems clear that the "shoelace formula" of Meister and Gauss* is needed, and a little program to apply that would be nice.

Peter

* <https://en.wikipedia.org/wiki/Shoelace_formula>

We could also seek the points {A, B, C} of the intersections of d1 and d2, d2 and d3, d1 and d3 (with solve for example) and then S=1/2.||AB /\ AC ||
(08-12-2015 09:29 PM)Gilles Wrote: [ -> ]then S=1/2.||AB /\ AC ||

Which of course is the same since:

\begin{align} (B-A)\wedge (C-A)&=B\wedge C-B\wedge A-A\wedge C+A\wedge A \\ &=B\wedge C+A\wedge B+C\wedge A+0 \\ &=A\wedge B+B\wedge C+C\wedge A \\ \end{align}

Cheers
Thomas
(08-12-2015 04:22 PM)Peter Murphy Wrote: [ -> ]The following problem, from the 2013 Houston calculator exam, seems conceptually simple to solve, but the manipulations necessary to solve it from start to finish on the 50g are not so obvious:

14. There are 3 points of intersection of pairs of the lines 2x+3y=4, 7x-2y=3 and
x+y=15. Give the area of the triangle determined by these three points.

It seems clear that the "shoelace formula" of Meister and Gauss* is needed, and a little program to apply that would be nice.

Peter

* <https://en.wikipedia.org/wiki/Shoelace_formula>

Are these questions expected to be solved solely via a calculator? Are these supposed to be approached via programming? Is some pencil to paper math allowed to simplify this? I worked an answer (A = 285 19/25 ?) by first rotating x+y=15 to y=15 then rotating the other lines slopes CCW by the same amount. Finding the intersection of these transformed lines yield the base, height of the triangle.
Only a numerical answer is required; solvers are free to rotate lines (which sounds like a very neat idea!) or to apply any other solving skills they may bring to bear.

(In 2013, problem 9 asked for the sum of the first 213 odd integers. One hopes that some contestants did this in seconds, on paper, with a trivial calculation at the end.)

But if we can become more competent users of our calculators (or if we can show others how they may do so) by putting them to work even when it's not required, we may want to do so, confirming results obtained more easily by thinking outside the calculator box. Or sleeve, as the case may be.

Peter
(08-12-2015 04:22 PM)Peter Murphy Wrote: [ -> ]The following problem, from the 2013 Houston calculator exam, seems conceptually simple to solve, but the manipulations necessary to solve it from start to finish on the 50g are not so obvious:

14. There are 3 points of intersection of pairs of the lines 2x+3y=4, 7x-2y=3 and
x+y=15. Give the area of the triangle determined by these three points.

I tried to calculate the area in a different way:
Code:
 [4 3] [[2 3][7 -2]] LSQ [4 15] [[2 3][1 1]] LSQ [3 15] [[7 -2][1 1]] LSQ 3 COL→ [1 1 1] 1 ROW+ DET 2 / ABS
S=250.88
(08-13-2015 01:10 AM)Peter Murphy Wrote: [ -> ](In 2013, problem 9 asked for the sum of the first 213 odd integers. One hopes that some contestants did this in seconds, on paper, with a trivial calculation at the end.)

Yes, just one trivial calculation. Gerson.
(08-13-2015 01:10 AM)Peter Murphy Wrote: [ -> ]Only a numerical answer is required; solvers are free to rotate lines (which sounds like a very neat idea!) or to apply any other solving skills they may bring to bear.

(In 2013, problem 9 asked for the sum of the first 213 odd integers. One hopes that some contestants did this in seconds, on paper, with a trivial calculation at the end.)

But if we can become more competent users of our calculators (or if we can show others how they may do so) by putting them to work even when it's not required, we may want to do so, confirming results obtained more easily by thinking outside the calculator box. Or sleeve, as the case may be.

Peter

Thanks and I agree that these are good exercises to explore. BTW - my idea to rotate the lines might work, but I did it wrong... so far.
Gravedigging, maybe, but this is a nice source of problems. Thanks for sharing!
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