There are actually a couple of threads on math forums of people being similarly confused, and one of the reasons I gave up was that I proved that 2x = 3x or similar a couple of times. I did wonder if there’s an inconsistency…
Right, or nx = 180 so my n = 12
5x + Zx = nx
6x + Yx = nx
and we know Zx = Yx + X so Zx is 1 larger than Yx.
Like you said we know the answer Zx = 7x, Yx = 6x, in my reverse engineering n =12x all of that yields X is 15, but would like to see it mathematically proved.
Yay, I never thought that my post would gather so much attention. This cheers me up.
However I want to clarify some points from reading all the replies.
This problem is not wrongly stated or anything. It is correct.
Yes, as it was indicated above, these questions are deliberately drawn in a way that is not on scale so you cannot measure it up right from the drawing.
If the word sexagesimal confuses you, disregard this. I was trying to indicate the usual degrees symbol as 15°. Base 60 or sexagesimal is better explained in the wikipedia entry.
Yes, the sum of the interior angles in a quadrilateral is given by, S=180(n-2) where n is the number of sides so it is 360. But this alone will not do much in this problem.
It wouldn’t be of much difference to upload a picture of the original problem from the book as it is some dated and it would not look clear. I also don’t have a camera available now. This is an exact replica of the original problem.
I think the solution for this without simmilarity requires a simple geometrical argument that uses some creative construction. As trying to make system of equations is just too much and clearly not the intended solution.
I don’t know should I contact Academia Sinica for this or Taiwan Mathematical Society, if they do reply who knows.
It is just the puzzling part of this problem that amuses me. Still I don’t know how to solve it.
What upsets me a bit is how this problem would never exist in real life.
Kites in real life would be 3D, you could just try to measure the angle with a compass if you were really interested, then when you realise that’s a right pain then you throw it away and buy yourself a passion fruit smoothie.
School is a total joke in the first place. Children learn abstract bullshit and then are expected to use those lessons in life.
We are a group of intelligent, kind and hard working people and we have no fucking clue what the answer to that ridiculous question is.
The real answer is: Go shove X up your ass, it’s a waste of my time trying to figure out the answer.
Theoretically the angles all “work,” but I decided to actually try to make the dang thing. I started with doing the basic angle sizes from the information in the example and basic triangle/line math:
To draw the shape I started with with line AB. Then I drew guidelines out at the correct angles based on whatever X I was using (5/10/20) from A & B. C & D were found where those guidelines met. Then I connected C & D. You can see from the pictures that doing it this way results in the correct expected angles in all of the places except the CD angles.
The bottom C angle is always correct but the top one never equals X. It is too big if my chosen X is less than the given correct answer of 15, or too small the other way. This correlates in some way with what @fonserbc was trying to say earlier I think.
I guess, we may live in non Euclidean space (some black hole sneaked in nearby?) or your desk is uneven?
Since x is arbitrary it should be possible to construct the kite:
draw parallel lines
select A and C arbitrarily
Construct D so that AD = DC
Adjust B to get angles right
I don’t know what you’re looking at but the three triangles I created have AD and BC as perfectly parallel. The first one isn’t because that was freehand just to copy the example. You can tell they are parallel because they are the same angle up from the base line of AB (3x).
Ok I have better explanation which should help with construction:
(construct using ruler and compass)
Draw parallel lines
Select A and C arbitrarily
Construct D so that AD = DC:
(1) construct bisector of AC (which will be a height of equilateral triangle ACD)
(2) bisector of AC will intersect line A at point D (we have constructed D)
Point B is totally independent of all the other points and will only change angle CAB = 2x (you can “move” pint B on line C to adjsut this angle until it is CAB = 2x = 2* DAC)
To find B, construct at A an arc and using compass, replicate angle DAC =x two times from AC.
The ray from A will intersect line C - this is point B
