to the cross product 
reveals that a negative determinant [of 
] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
and 
,
and 
although its supposed to be a quick comparison.
chance of landing heads when flipped. Krithik is playing a game where he starts with 
point. Every turn, he flips the coin, and if it lands heads, he gains 
points. Find the expected number of turns until Krithik wins.
be an isosceles triangle with 
Let 
be a point on the circumcircle of 
Let 
intersect the extension of 
at 
Let 
be the midpoint of segment 
and let 
be the intersection of 
and 
Let the extension of 
intersect 
and the circumcircle of 
and 
respectively. Given that 
and 
compute the length of 
.
be a subset of 
such that no sum of three distinct elements of 
.
.
that satisfy the equation 
.
be a circle, and let a line 
intersect 
and 
Circles 
and 
are internally tangent to 
and 
respectively, and both are tangent to 
Similarly, circles 
and 
are externally tangent to 
and 
respectively.
the segment 
has a length of 
and 
find the length of segment 
and 
Prove that
.
Prove that
and 
Prove that
Y by
(i) Prove that no 
diagonals of a regular heptagon (
-sided polygon) are concurrent at a point other than a vertex of the heptagon. A diagonal is a line connecting two non-adjacent vertices.
(ii) How many points of intersection of pairs of diagonals lie within the heptagon?
(iii) Into how many compartments is the heptagon dissected by the diagonals?
(iv) Assuming that for
odd no diagonals of a regular -gon are concurrent, generalize (ii) and (iii) for the regular -gon.
diagonals of a regular heptagon (
-sided polygon) are concurrent at a point other than a vertex of the heptagon. A diagonal is a line connecting two non-adjacent vertices.(ii) How many points of intersection of pairs of diagonals lie within the heptagon?
(iii) Into how many compartments is the heptagon dissected by the diagonals?
(iv) Assuming that for
odd no diagonals of a regular -gon are concurrent, generalize (ii) and (iii) for the regular -gon.This post has been edited 2 times. Last edited by parmenides51, Jan 23, 2021, 9:26 AM
![[asy]
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label("F",(-3,10));
label("G",(7,8));
draw((10,0)--(-2,-10),red);
draw((6,-8)--(-9,4),green);
[/asy]](http://latex.artofproblemsolving.com/6/c/0/6c0c197cd35a486f8e2e637310f87caa62190eb8.png)
![[asy]
draw((10,0)--(6,-8)--(-2,-10)--(-9,-4)--(-9,4)--(-2,10)--(6,8)--cycle);
label("A",(11,0));
label("B",(6.75,-8));
label("C",(-2.5,-11));
label("D",(-10,-5));
label("E",(-10,4));
label("F",(-3,10));
label("G",(7,8));
draw((10,0)--(-2,-10),red);
draw((6,-8)--(-9,-4),red);
[/asy]](http://latex.artofproblemsolving.com/4/f/9/4f9ca2441b23b141e1ed1fe7e31d058959889a87.png)
![[asy]
draw((10,0)--(6,-8)--(-2,-10)--(-9,-4)--(-9,4)--(-2,10)--(6,8)--cycle);
draw((10,0)--(-9,-4),green);
draw((-2,-10)--(6,8),green);
label("A",(11,0));
label("B",(6.75,-8));
label("C",(-2.5,-11));
label("D",(-10,-5));
label("E",(-10,4));
label("F",(-3,10));
label("G",(7,8));
[/asy]](http://latex.artofproblemsolving.com/a/a/c/aacf332d45e7edf363dad164a1389c852bcaf11e.png)
![[asy]
draw((10,0)--(6,-8)--(-2,-10)--(-9,-4)--(-9,4)--(-2,10)--(6,8)--cycle);
draw((10,0)--(-9,4),green);
draw((6,8)--(-2,-10),green);
label("A",(11,0));
label("B",(6.75,-8));
label("C",(-2.5,-11));
label("D",(-10,-5));
label("E",(-10,4));
label("F",(-3,10));
label("G",(7,8));
[/asy]](http://latex.artofproblemsolving.com/a/5/b/a5bdbabf84cf5d32cba09dbc3fa11f0d4d64b4ce.png)
and 
as 
.
intersected the two other diagonals at a common point which must be the midpoint. Call the midpoint 
.![[asy]
draw((10,0)--(6,-8)--(-2,10)--(6,8)--cycle);
label("A",(11,0));
label("B",(6.75,-8));
label("F",(-3,10));
label("G",(7,8));
label("M",(1,0));
draw((10,0)--(2,1)--(6,8),green);
[/asy]](http://latex.artofproblemsolving.com/6/7/2/672297cbdfe7bd5cab4cfc63f99824884cbabbdf.png)
,
.
and the base of the altitude from 
,
.
,
intersections of diagonals.
because for every four vertices of the heptagon we select, we can create exactly one pair of intersecting diagonals that intersect. And from part one, no three diagonals intersect at one point. The other way around is easy to see. For every pair of intersecting diagonals, it corresponds to exactly four points. Thus we have a one to one correspondence, so the answer is 
