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?
Correct me if i´m wrong but doesn´t the fact that it´s rotating means it cannot stay on the same line?
Chase Jenkins
Agreed
Jacob Long
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.
Nathaniel Lee
ABC is isosceles isn't it rotation just describes how the first picture evolves in the second I think
Jose Gutierrez
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.
Caleb Harris
>ABC is isosceles isn't it Yes it is, forgot to mention that too. AB = AC is what it says in the description.
Landon Wilson
Then angle BAD = angle CAN, because MN and BC are paralell, right?
Ryder Morgan
BAD and BDE are always 90deg user, they have the little square inside
Brody Taylor
>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.
Brayden Brooks
That's right, but only when D is on the left side of A. I think you really ment ∠BAM = ∠CAN.
Zachary Thompson
Yeah, that would be better, sorry, got carried away looking at the drawing.
Aiden Kelly
holy shit, I have no clue of how even to solve it.
Jayden Martin
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?
Alexander Moore
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
Kevin Murphy
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.
Brandon Foster
Tried chasing angles... didn't get it done.
Jaxson Brown
Does this approach look good
Xavier Anderson
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.
Parker Edwards
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
Ayden Brooks
looking at it again, the passage for O isn't a general case, so no
Tyler Brooks
Protip: draw a circle with diameter BP, then A and D lie on the circumference.
Ryan Rogers
Then BPAD is inscribed in a circle, opposte angles sum up to 180°or whatever...?
David Lewis
This is p difficult. I can't even read the question.
Samuel Clark
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|
Asher Ross
Take your terrorist language back to .
Christopher Lee
Noice, but I guess one should solve it with geometric means...
Xavier Wright
Shieet this boi is hard af.
Kevin Perez
Solving it geometrically is probably simpler, one you figure it out.
Wyatt Ward
^ DAP = 45. As ADBP is a cyclic quadrilateral, DBP = 45. Hence, triangle DPB is isosceles.
Jonathan Myers
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
Jaxon Perry
>DAP = 45 ???
Josiah Gutierrez
He's referring to 2nd and 3rd figures. For 1st it's DAB and DPB
Hunter Howard
I don't see what DAB = 45 in the first figure either.
Charles Stewart
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?
Mason Gutierrez
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.
Samuel Long
So is there a geometric solution in this thread? I can't see one.
Levi Reed
No I don't think so. is closest but that's just in one case. So still pretty far off.
Jace Cook
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.
Connor Mitchell
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.
Leo Thompson
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.
Juan Bailey
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?
Jace Ortiz
>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.
Isaiah Smith
Nah, nevermind. I realized that the position of E is irrelevant, as long as it's on the DE line.
Noah Turner
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.
Alexander King
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
Cameron Taylor
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.
Bentley Clark
>inscribed angle theorem Damn, can't believe I missed that. Yea, it proves all 3 cases quite elegantly.
Benjamin Scott
How do you know that ADBP is a cyclic quadrilateral?
Landon Rogers
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.