be a positive integer. A Japanese triangle consists of 
circles arranged in an equilateral triangular shape such that for each 
,
,
,
row contains exactly 
circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of 
,
such that in each Japanese triangle there is a ninja path containing at least 
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
Y by Adventure10
Ada drew a triangle, chose a point on each side and chose a point 
on the inside of the triangle. Then, drew segments that join with the other six points (the three vertices and the three points that are on the sides). In this way the initial triangle was divided into six isosceles triangles. Show, by example, how Ada could have achieved this.
on the inside of the triangle. Then, drew segments that join with the other six points (the three vertices and the three points that are on the sides). In this way the initial triangle was divided into six isosceles triangles. Show, by example, how Ada could have achieved this.This post has been edited 1 time. Last edited by parmenides51, May 14, 2019, 12:55 AM
Reason: latex
Reason: latex
be an equilateral triangle and let 
be an isosceles right triangle. Then let 
be the midpoint of 
so 
and 
are isosceles. Construct point 
on 
such that 
and 
,
on 
.![[asy]
pair M, B, C, A, P, D, E;
M = (0,0); B = (-(sqrt(3)+1)/2, 0); C =((sqrt(3)+1)/2, 0); A = (0, (3+sqrt(3))/2); P = (0, (sqrt(3)+1)/2); D = (sqrt(3)/2, sqrt(3)/2); E = (-sqrt(3)/2, sqrt(3)/2);
draw(A--B--C--cycle); draw(A--P, heavycyan); draw(P--D, heavycyan); draw(P--E, heavycyan); draw(D--C, heavycyan); draw(E--B, heavycyan); draw(P--B); draw(P--C); draw(P--M, heavygreen); draw(B--C, heavygreen);
dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A$", A, dir(A)); dot("$P$", P, NE); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$M$", M, S);
[/asy]](http://latex.artofproblemsolving.com/2/4/4/244fa5aab32cafc18773f7c55d57f98fc8ef4d07.png)
