Chinese 9th Grade Math

>>8187570
>geometry
>9th grade
Americans...

>>8187570
Correct me if i´m wrong but doesn´t the fact that it´s rotating means it cannot stay on the same line?

>>8187713
Agreed

>>8187713
Well it can rotate to figure 1 and 2 and be on the same line obviously, but i've got no clue whats going on in 3, figure 3's BDE is clearly not the same as figure 1 or 2's BDE. Fucking chinks.

>>8187570
ABC is isosceles isn't it
>>8187713
rotation just describes how the first picture evolves in the second I think

>>8187713
>>8187730
>>8187738
Forgot to mention the triangle vary in size.
You should prove that the BD = DP for all cases when D is on the MN line, no just the three cases given.

>>8187747
>ABC is isosceles isn't it
Yes it is, forgot to mention that too.
AB = AC is what it says in the description.

>>8187775
Then angle BAD = angle CAN, because MN and BC are paralell, right?

>>8187780
BAD and BDE are always 90deg anon, they have the little square inside

>>8187782
>BAD and BDE are always 90deg anon, they have the little square inside
You are thinking of BAC, anon, BAD is not given and is an exterior angle to BAC., also, you didn't mention CAN.

>>8187780
That's right, but only when D is on the left side of A.
I think you really ment ∠BAM = ∠CAN.

>>8187787
Yeah, that would be better, sorry, got carried away looking at the drawing.

holy shit, I have no clue of how even to solve it.

>>8187801
Im thinking that by proving ∠DBP = ∠BPD it will be solved, but then comes the question, how can we determain even one of these angles?

i know that i can do it, but i would go down a long winding road expressing all points and line equations in 2D space

>>8187806
You're absolutely right. And if we determine one we have the other as well, since we know that ∠BDP = 90°.
>In other words: prove that ∠DBP = 45°, or that ∠BPD = 45°, and we're done.

File: ChanBDequalsDP.gif (126KB, 737x474px) Image search: [Google]

126KB, 737x474px

Tried chasing angles... didn't get it done.

File: IMG_20160706_214338865.jpg (3MB, 2340x4160px) Image search: [Google]

3MB, 2340x4160px

Does this approach look good

>>8187817
>>8187867
When D is directly above B, P intersects the AC tangent exactly on A.
That means BPD = BAM = 45°, which means BD = DP.

Don't know how to prove it for every other case though.

>>8187872
it would show that if you fix one between b and d BA=AD for a particular d or b (the one you didn't fix), I don't know that would be a general demonstration though

>>8187872
>>8187879
looking at it again, the passage for O isn't a general case, so no

Protip: draw a circle with diameter BP, then A and D lie on the circumference.

>>8187889
Then BPAD is inscribed in a circle, opposte angles sum up to 180°or whatever...?

This is p difficult.
I can't even read the question.

Algebra works pretty well:

Scale and rotate so:
B=(0,0)
C=(2,0)
A=(1,1)
Let D=(d,1)
|BD|=sqrt(1+d^2)
Line DE, parametric form: (d,1)+s*(1,-d)
Line AC: y=2-x
At intesection point P:
1-d*s = 2 - (d+s), so s=1, so
P=(d+1,1-d) and
|DP|=sqrt(1+d^2)=|BD|

>>8187934
Take your terrorist language back to >>>/pol/.

>>8187934
Noice, but I guess one should solve it with geometric means...

Shieet this boi is hard af.

>>8188021
Solving it geometrically is probably simpler, one you figure it out.

>>8187889
^
DAP = 45. As ADBP is a cyclic quadrilateral, DBP = 45. Hence, triangle DPB is isosceles.

well BD is a set of line and if you add (-y, x) if a vector/point on the line is (x, y) to each point on the line you obviously get another line which is the line of all points P. So they lie on a line
thid line is the line through AC. There's no reason for the second triangle, they literally drew a triangle out of the line formed by the different P and some other arbitrary static points.
so:
d = (hx, hy) + r* (kx, ky) (h and k are some arbitrary vectors)
p = d + (p - d)
p = (hx, hy) + r* (kx, ky)
+ (-hy, hx) + r* (-ky, kx)
=(hx - hy, hy + hx) + r*(kx-ky, ky+kx)
so all p lie on a line. since no information is given about A and C I dont see anything else to prove

>>8189165
>DAP = 45
???

>>8189316
He's referring to 2nd and 3rd figures. For 1st it's DAB and DPB

>>8189320
I don't see what DAB = 45 in the first figure either.

>>8187570
I'm kinda confused as to how E is determined. Is BDE defined by BD=DP? That would make this problem sort of redundant, no?

Maybe I'm terribly stupid, but this could actually be a ninth-grader's problem. Would rotating BDE so that D=A prove that BD must equal DP?

>>8189353
You can disregard E completely, it's an arbitrary point. It's either there for confusion or simplification, depending or who reads it.
>Would rotating BDE so that D=A prove that BD must equal DP?
That's actually the one case when BD=/=DP. P is defined as when DE intersects the AC tangent.
When D=A all of DE is inside that tangent, and thus P is not properly defined.

So is there a geometric solution in this thread? I can't see one.

>>8190056
No I don't think so. >>8187875 is closest but that's just in one case. So still pretty far off.

Does this break down if triangles ABC and DBE are the same? It's impossible to provide a general answer for this unless specific cases are excluded than.

>>8187570
isnt this just saying that when point D is at the same spot as point A, then BDE is the exact same triangle as ABC, therefore proving BD is = to DP

since the triangle is able to rotate and D must stay on MN that means the triangle must be able to vary in size.

>>8191028
It doesn't break down. You can see here:
>>8187934
that it doesn't depend on what d is, d can be any real number. Just consider extending the line AC to all of y=2-x.

>>8191374
sorry, I was wrong. when d=1, s can be anything and P can be any point on the line through AC. It's unique though for [math]d\ne 1.[/math]

Is there a way to convert >>8187934
into a geometric proof?

>>8187570
>BC is just a line that extends to infinity, and C is the intersection point created by the angle AC and BC
So you know where C is in space at all times.

The problem is that because there's no clause stating that the angle DE is identical to the angle AC, you can't guarantee anything about the position of E in space, other than that it's "at a 90 degree angle to B", which is useless.

I see in the chink that they're saying something about DE and CA, please translate properly, OP.

>>8191662
Nah, nevermind. I realized that the position of E is irrelevant, as long as it's on the DE line.

Attempting to try to explain it in an intuitive manner, because I'm terrible at math:

ABC is a triangle where all the angles are known (right triangle, isosceles, which can be derived thanks to MN||BC)

Now, ignore the BE line. It's basically totally irrelevant.

What you want is the line BP (not drawn).
Somehow you need to prove that the angles DBP and BPD are equal.

>>8189316
Whoops. Meant DAB in the first figure and DAP in the third. I also just noticed that the proof doesn't hold for the 2nd figure. Maybe you could try reflecting triangle DPB about line BP...

>>8190056
See this
>>8189165

>>8191686
Indeed, and you could use cyclic quadrilaterals to prove that

>>8191706

I see! Thank you very much.

I think you could also use the inscribed angle theorem. Angles DAB and DPB are inscribed in a circle and equal, as they cut the same chord DB.

>>8191968
>inscribed angle theorem
Damn, can't believe I missed that. Yea, it proves all 3 cases quite elegantly.

>>8187889
>>8189165
>>8191706
How do you know that ADBP is a cyclic quadrilateral?

>>8192036
BDP and BAP are two right angled triangles which share the same hypotenuse BP, so just draw a circle with radius = BP/2 centered at the mid-point of BP. A, D, B, P all lie on this circle as the midpoint of the hypotenuse is equidistant from all vertices. Hence, ABDP is cyclic.

http://mathworld.wolfram.com/RightTriangle.html

(of course this assumes all 4 points to be distinct, but if they aren't, the proof is almost trivial)