More Harmonies

by agbdmrbirdyface, Aug 10, 2016, 10:28 PM

Oops here's the geo problem:
110 Geometry Problems #5 wrote:
Let $C$ be a circle and let $P$ be a point outside of $C$. The tangents from $P$ intersect the circle at $A$ and $B$, respectively. Let $M$ be the midpoint of segment $AP$ and $N$ the second intersection of the line $BM$ with circle $C$. Prove that $PN$ = $2MN$.

Solution:
This post has been edited 1 time. Last edited by shiningsunnyday, Aug 11, 2016, 8:22 PM

Comment

5 Comments

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Wait this looks nice lemme try it.

by shiningsunnyday, Aug 10, 2016, 11:01 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Alternative sol

by shiningsunnyday, Aug 10, 2016, 11:16 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Wait darn that's OP
There's another solution with a quick Law of Sines bash, but it's not great.
This was actually a problem from one of Virgil Nircea's posts, I think.

Right... I totally forgot about Cosmin's AoPS-user-recommendations. :maybe:
This post has been edited 1 time. Last edited by shiningsunnyday, Aug 11, 2016, 11:24 AM

by agbdmrbirdyface, Aug 11, 2016, 3:47 AM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
[asy]
size(200);
defaultpen(fontsize(10pt));
pair A, B, C, P, M, NN;
C = (0,0);
P = (3,0);
path circ = Circle(C, 1);
A = tangent(P, C, 1, 1);
B = tangent(P, C, 1, 2);
draw(circ);
draw(A--P--B--cycle);
M = (A+P)/2;
draw(B--M);
pair [] N = IPs(B--M, circ);
pair [] X = IPs(P--(2*N[1]-P), circ);
draw(P--X[1]);
NN = rotate(180, M)*N[1];
draw(N[1]--A--NN--P, red);
draw(M--NN, red);
draw(P--M--B--P--N[1]--M, green);
dot("$N'$", NN, dir(90));
dot("$A$", A, dir(70));
dot("$B$", B, dir(290));
dot("$C$", C, SW);
dot("$P$", P, dir(0));
dot("$X$", X[1], dir(200));
dot("$M$", M, dir(70));
dot("$N$", N[1], dir(330));
[/asy]

Diagram mostly thanks to pinetree1

I did the colors: red is my solution, green is shiningsunnyday's solution

Nice, thanks!!!
This post has been edited 1 time. Last edited by shiningsunnyday, Aug 11, 2016, 8:22 PM

by agbdmrbirdyface, Aug 11, 2016, 7:56 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Oh so this is what you used the diagram for...

by pinetree1, Aug 12, 2016, 3:07 PM

To share with readers my favorite problem I came across today :) (Shout for contrib.)

avatar

shiningsunnyday
Archives
- May 2017
Shouts
Submit
  • this guy is an absolute legend. much love wherever you are Michael

    by LeonidasTheConquerer, Aug 5, 2024, 9:37 PM

  • amazing blog

    by anurag27826, Jun 17, 2023, 7:20 AM

  • hi i randomly found this

    by purplepenguin2, Mar 1, 2023, 8:43 AM

  • can i be a contributor please?

    by cinnamon_e, Mar 10, 2022, 6:58 PM

  • orzorzorzorzorzorozo

    by samrocksnature, Jul 16, 2021, 8:25 PM

  • 2021 post

    by the_mathmagician, May 5, 2021, 3:28 PM

  • Let $ ABC$ be an equilateral triangle of side length $ 1$. Let $ D$ be the point such that $ C$ is the midpoint of $ BD$, and let $ I$ be the incenter of triangle $ ACD$. Let $ E$ be the point on line $ AB$ such that $ DE$ and $ BI$ are perpendicular. $ \

    by ARay10, Aug 25, 2020, 5:55 PM

  • Nice blog! :)

    by User526797, Jan 12, 2020, 4:48 PM

  • oh my gosh it's been so longggggg.... contrib? what does that mean?

    by adiarasel, Dec 1, 2019, 8:31 PM

  • 2019 post

    by piphi, Aug 10, 2019, 6:32 AM

  • hi contrib please

    by Emathmaster, Dec 27, 2018, 5:38 PM

  • hihihihihi contrib plzzzzz

    by haha0201, Aug 20, 2018, 3:58 PM

  • contrib please

    by Max0815, Aug 1, 2018, 12:35 AM

  • contrib /charmander

    by mathmaster2000, Apr 16, 2017, 4:59 PM

  • for contrib

    by SomethingNeutral, Mar 30, 2017, 7:57 PM

270 shouts
Tags
About Owner
  • Posts: 1350
  • Joined: Dec 19, 2014
Blog Stats
  • Blog created: Jun 11, 2016
  • Total entries: 193
  • Total visits: 30957
  • Total comments: 579
Search Blog
a