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Square
Posted: July 31st, 2021, 1:36 pm
by cinelli
A----------------------------------------B
|. . |
| .45 . |
| . . |
| . . |
| . . |
| . . |
| . . 70 |
| . E
| . .|
| . . |
| . . |
| . . |
| . . |
| . . |
| . . |
| . . |
D----------------F-----------------------C
ABCD is a square of side length 1 unit. A straight line is drawn from A to meet BC at E so that angle AEB is 70 degrees. Another straight line is drawn from A to meet CD at F so that angle EAF is 45 degrees.
Using only elementary geometry (a) what is angle AFE? and (b) what is the perimeter of triangle CEF?
Cinelli
Re: Square
Posted: July 31st, 2021, 7:52 pm
by UncleEbenezer
Easy to misread what you're asking there. Elementary geometry is too easy, and the answers are suspiciously round numbers. Did you mean to say no calculators or computers? I've only solved it with the aid of trig functions, but let's spoiler-protect it anyway.
Lots of right-angled triangles and unit sides to work with.
Looking at triangle BAE, angle BAE is 20° and side BE is tan(20). Similarly angle FAD is 25° and side DF is tan(25). From this,
FC = 1 - tan(25)
EC = 1 - tan(20)
FE = sqrt(FC^2+EC^2). Expanding that doesn't look particularly exciting, so I won't.
Adding those up gives the perimeter. But that's surely too easy for a Cinelli problem?
How about that angle? Consider G, vertically above F on AB. Then the angle is AFG + GFE, and we already have AFG=25°.
GFE is of course 90 - EFC, or equal to FEC. And FEC = arc tan (FC/EC), giving AFE = AFG+GFE = 25+atan(FC/EC)
Again too easy for a Cinelli problem. Evaluating leads to suspiciously round numbers (65 degrees and exactly 2) hinting at the existence of a more elegant solution. And AE is Valentines Day: it bisects FEB. 
Corollary: my initial reading of the 70° as angle AEF is coincidentally correct!
That also finds me a pattern which is clearly at the heart of your solution. But I think I've spoiled that for myself.
Re: Square
Posted: August 1st, 2021, 3:14 pm
by cinelli
UncleEbenezer wrote:Easy to misread what you're asking there. Elementary geometry is too easy, and the answers are suspiciously round numbers. Did you mean to say no calculators or computers?
No trig, no calculators, no computers. Just geometry.
Cinelli
Re: Square
Posted: August 2nd, 2021, 2:48 pm
by nmdhqbc
mindyourdecisions
Re: Square
Posted: August 3rd, 2021, 11:09 am
by Gengulphus
cinelli wrote:A----------------------------------------B
|. . |
| .45 . |
| . . |
| . . |
| . . |
| . . |
| . . 70 |
| . E
| . .|
| . . |
| . . |
| . . |
| . . |
| . . |
| . . |
| . . |
D----------------F-----------------------C
ABCD is a square of side length 1 unit. A straight line is drawn from A to meet BC at E so that angle AEB is 70 degrees. Another straight line is drawn from A to meet CD at F so that angle EAF is 45 degrees.
Using only elementary geometry (a) what is angle AFE? and (b) what is the perimeter of triangle CEF?
Spoiler...
I know that 70 degrees doesn't have any very special geometric properties, so I'll generalise the problem to say the angle AEB is x+45 degrees, where x lies between 0 and 45 (to ensure that the first straight line drawn from A does indeed meet the line segment BC).
By using the fact that the three angles of a triangle add up 180 degrees and that the three angles BAE, EAF and FAD add up to the angle BAD, which is 90 degrees, it is easy to deduce that angle BAE = 45-x degrees, angle FAD = x degrees and angle AFD = 90-x degrees.
Reflect the triangle ABE in its side AE to produce another triangle AGE, and reflect the triangle ADF in its side AF to produce another triangle AHF. Triangles ABE and AGE are similar, so angle BAG is twice angle BAE, i.e. 90-2x degrees, the distance AG is 1 unit and angle AGE is 90 degrees. Similarly, triangles ADF and AHF are similar, so angle HAD is twice angle FAD, i.e. 2x degrees, the distance AH is 1 unit and angle AHF is 90 degrees.
From those, we see that G and H are both the same distance away from A, and in the same direction, so they are the same point. Furthermore, that implies that the angles AGE and AGF (= angle AHF) add up to 180 degrees, so the line segments EG and FG are collinear - i.e. together they make up the line segment EF, and the point G=H lies on that line segment. So again by the similarity of the reflected triangles, angle AFE = angle AFG = angle AFD = 90-x degrees (i.e. 65 degrees for the specific problem posed), and perimeter(CEF) = length(CF) + length(FG) + length(GE) + length(EC) = length(CF) + length(FD) + length(BE) + length(EC) = length(CD) + length(BC) = 2 units.
A----------------------------------------B
|.\ . 45-x 90 |
| . \ . |
| . \ 45-x . |
| x . x \ . |
| . \ . |
| . \ . |
| . \ . x+45|
| . \ E
| . \ x+45 .|
| . \ . |
| . \ . |
| . \ 90 . |
| . 90 G/H |
| . . |
| . 90-x . |
| 90 90-x . . |
D----------------F-----------------------C
Gengulphus
Re: Square
Posted: August 4th, 2021, 3:58 pm
by Gengulphus
A more succinct version of the above spoiler...
Take a square piece of paper and fold corner B to somewhere inside the square, with the crease passing through corner A. Fold corner D similarly to the same point inside the square, and look at what the edges BC and CD have become.
Might be criticised for using origami rather than elementary geometry, though! ;-)
Gengulphus
Re: Square
Posted: August 4th, 2021, 4:34 pm
by 9873210
Gengulphus wrote:A more succinct version of the above spoiler...
Take a square piece of paper and fold corner B to somewhere inside the square, with the crease passing through corner A. Fold corner D similarly to the same point inside the square, and look at what the edges BC and CD have become.
Might be criticised for using origami rather than elementary geometry, though! ;-)
Gengulphus
Might be criticized for proving the wrong thing too. While it's a powerful hint what it actually shows is B => A rather than A => B (see spoiler for A and B) which is required to solve the puzzle. In this case A <==> B so both are true, but they need to be proved separately (or you need to show that each step in the proof is reversible)
*
In this case A is that "EAF is 45 degrees" and B is that G and H are the same point.
Re: Square
Posted: August 5th, 2021, 8:10 pm
by cinelli
Reply to Gengulphus’s solution.
This solution is ingenious. As nmdhqbc has hinted, this is a puzzle set and solved by Presh Talwalkar and the solution he gives is different. I must say though that I prefer the origami method. Presh Talwalkar gives an alternative construction:
. A----------------------------------------B
.|. . 20 |
. | .45 . |
. | . . |
.20 |25 . . |
. | . . |
. | . . |
. | . . 70 |
. | . E
. | . .|
. | . . |
. | . . |
. | . . |
. | . . |
. | . . |
. | . . |
. 70 | 65 . . |
G----------------D----------------F-----------------------C
Rotate triangle ABE clockwise by 90 degrees about A to create triangle ADG. Angle BAE is 20 deg so DAG = 20 deg, DAF = 25 deg, AFD = 65 deg, AGD = 70 deg. FAG = 20 + 25 = 45 deg. Then triangles AFE and AFG are congruent SAS. (a) So AFE = AFG = 65 deg. (b) Let BE = x, DF = y. Then CE = 1-x, CF = 1-y, EF = FG = x+y. Perimeter = (1-x) + (1-y) + (x+y) = 2 units. And as Gengulphus has pointed out, there is nothing special about angle AEB being 70 degrees. The triangle’s perimeter is 2 for any value between 45 deg and 90 deg. Cinelli