Chinese 9th Grade Math

Saw this math challenge some time ago. Supposedly it's for 9th graders in China, but I doubt it. In any case I can't solve it, and it is driving me nuts.

>The challenge is to prove that the distance BD is equal to the distance DP.
B is a static point.
BC is parallell with MN.
BDE is a triangle that rotates around B, and D is always on the MN line.
BDE's rotation is displayed in the three figures.
P is the point where DE intersects with the AC "tangent".

Hope I've been clear enough. Tell me if not, and I'll try to explain better.
Any geniuses here that can help me?

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>geometry
>9th grade
Americans...

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

Agreed

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.

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

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.

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

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

BAD and BDE are always 90deg user, they have the little square inside

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

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

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.

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

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.

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

Does this approach look good

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.

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

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.

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|

Take your terrorist language back to .

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

Shieet this boi is hard af.

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

^
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

>DAP = 45
???

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

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

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?

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.

No I don't think so. 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.

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.

It doesn't break down. You can see here:
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.

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 into a geometric proof?

>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.

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.

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...

See this
Indeed, and you could use cyclic quadrilaterals to prove that

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.

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

How do you know that ADBP is a cyclic quadrilateral?

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.

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)