AoPS Academy
+3 w
be a convex pentagon with 
and 
For each point 
in the plane, define 
The least possible value of 
can be expressed as 
where 
and 
are positive integers and 
is not divisible by the square of any prime. Find 
and 
,
and 
Adding, 
, so 
, with equality iff the aforementioned quadrilaterals are cyclic -- but, by 
angles, this means is the Fermat point of 
. Since that point minimizes 
, it must be the minimum.
.
be a triangle with 
, 
, and 
. Let 
be such that 
and 
, and define 
similarly.
denote the Fermat point of . This point satisfies 
, and it is well known that it minimizes 
(proof: smoothing). The main claim is the following. lies on 
, and thus also minimizes 
. Note that 
and 
are cyclic due to the 
and 
angles. Thus,
and similarly for 
so 
are collinear.
be the point on the other side of 
as 
such that 
is equilateral. Let 
However, by Ptolemy on 
,
By Ptolemy on 
and ,
which means the answer is
We know that 
, so coordinate bash to find that 
, so the answer is 
.
be the Fermat point of 
. Then, 
, so lies on . Therefore, we need to find 
. Since 
, the point 
such that 
is an equilateral triangle satisfies 
, so 
. By Ptolemy, 
and 
, so 
, implying 
. Therefore, the minimum is 
.
be the Fermat Point of , lies on . By definition, we have that 
. Therefore, 
and 
are cyclic so 
.
. Now, we find . We can use law of cosines to find 
, so 
. Thus, by law of cosines again 
. by constructing an equilateral triangle 
with and on opposite sides of . Note that 
passes through and furthermore 
is cyclic. By Ptolemy, 
. To find this length, we can drop the feet from and to to get 
and 
. 
has length 
and is the midpoint of . We already found the cosine, so it is easy to see that 
, so we can find the area of and divide by half of to find that the altitude from is 
. Then, 
so 
(I spent like 15 minutes trying to figure out why my answer was wrong; apparently 24-9 is not 13). Finally, we have 
by Pythag.
.
and 
. First note that these two triangles are right, and that is the Fermat Point of . Therefore this point indeed minimizes . It remains to prove it also minimizes 
. Thus the next claim is that 
are collinear. Note that both 
and 
are right angles, and our claim is proven. The rest of the problem now proceeds using 
methods. First we use cosine addition on 
to use LOC to find , and use law of Cosines on the 
triangles formed by the Fermat point, also equating sine area formula, we can solve for the sum of these lengths as well to be 
, by Ptolemy inequality we have 
, while the inequality is reached when both 
and are concyclic. Since 
, so lie on the same line. Thus, the desired value is then 
.
by LOC< the answer is then 
as the foot from to , which we claim is the optimal point. Note that we have minimized , and the cyclicities of and 
gives 
, making the Fermat point of and thus minimizing too.
. Notice we have the spiral similarity 
, and gives![\[f(X) = (BX+EX)+(AX+CX+DX) = BE+CF = BE\left(1+\frac{\sqrt 3}{2}\right) = \boxed{38+19\sqrt3}.\]](http://latex.artofproblemsolving.com/0/5/f/05fad6dbf2865d11adc439c53ce5654122933085.png)
![[asy]
size(300);
pair A, B, C, D, E, F, X;
A = (0,10);
C = (-4,0);
D = (11,0);
B = C + sqrt(3)/3*(A-C)*dir(90);
E = D + sqrt(3)/3*(A-D)*dir(-90);
F = A + (D-A)*dir(60);
X = foot(A, B, E);
draw(A--B--C--D--E--cycle^^C--A--D^^A--F);
draw(B--E^^C--F^^A--X^^D--X, dashed);
draw(circumcircle(A, B, C)^^circumcircle(A, D, E), blue);
dot("$A$", A, 2*dir(105));
dot("$B$", B, dir(225));
dot("$C$", C, dir(250));
dot("$D$", D, dir(290));
dot("$E$", E, dir(0));
dot("$F$", F, dir(45));
dot("$X$", X, 2*dir(270));
label("$14$", A--B, dir(135));
label("$7 \sqrt 3$", A--C, .1*dir(150));
label("$26$", A--E, dir(75));
label("$13 \sqrt 3$", A--D, .01*dir(45));
label("$24$", C--D, dir(270));
[/asy]](http://latex.artofproblemsolving.com/9/0/6/9062e17686530a3c938628df76a5908b4216ed63.png)
and 
are both 30-60-90 triangles, and 
and 
.
be the fermat point of . Then is minimized for 
. Moreover, since 
, we find that 
and 
are cyclic. Thus, 
, so lies on 
and thus is minimized for .
, and 
. By the law of cosines on 
, 
, and 
, we find that
Summing all three equations yields 
. Additionally, from computing the area of in two ways, we find that 
, so 
. We then find that 
, so 
.
. From Ptolemy's theorem on 
, we have 
, so 
. Similarly, 
, so 
. Thus, 
.
and showed that it was on by inversion but due to a massive skissue was not able to find 
after that.
, 
rfwbuh
extensions which makes it much more obvious that this problem is about the triangle , not some random pentagon.
![[asy]
import olympiad;
pair D=(0,0), C=(24,0), A=(39/2, 13sqrt(3)/2), E=(-13/2, 13sqrt(3)/2), B=(24+13/2, 3sqrt(3)/2);
draw(A--B--C--D--E--A);
draw(D--A--C);
label("$A$", A, N);
label("$B$", B, dir(1));
label("$C$", C, SE);
label("$D$", D, SW);
label("$E$", E, W);
draw(rightanglemark(A,D,E,32));
draw(rightanglemark(A,C,B, 32));
[/asy]](http://latex.artofproblemsolving.com/8/c/6/8c69338aabf97cf90ccf1490ec2feb624acbbbb5.png)
and 
are similar to 
-
Thus, 
![[asy]
import olympiad;
pair D=(0,0), C=(24,0), A=(39/2, 13sqrt(3)/2), E=(-13/2, 13sqrt(3)/2), B=(24+13/2, 3sqrt(3)/2);
draw(A--B--C--D--E--A);
draw(D--A--C);
label("$A$", A, N);
label("$B$", B, dir(1));
label("$C$", C, SE);
label("$D$", D, SW);
label("$E$", E, W);
draw(rightanglemark(A,D,E,32));
draw(rightanglemark(A,C,B, 32));
pair P=foot(A, B, E);
draw(B--E);
draw(D--P--A--P--C);
draw(rightanglemark(E,P, A, 32));
draw(rightanglemark(B,P, A, 32));
label("$P$", P, dir(260));
[/asy]](http://latex.artofproblemsolving.com/1/4/8/1480b4e5f1e3e0a3957ba6893a27cdda94634231.png)
and 
so 
and 
are cyclic.
and 
so 
as desired.
so 
We just need to evaluate 
now.
so 
so 
Now, 
so 
As such, by the law of cosines, 
so 
It now suffices to find 
and 
we have 
and 
so 
Also, ![$[ACD]=\dfrac{1}{2}\cdot AD\cdot AC\sin(\angle CAD)=\dfrac{1}{2}\cdot 13\sqrt{3}\cdot 7\sqrt{3}\cdot \dfrac{4\sqrt{3}}{7}=78\sqrt{3},$](http://latex.artofproblemsolving.com/9/f/3/9f37ccf668e8fe5edcc6e4666ebb722091e2e957.png)
and ![$78\sqrt{3}=[ACD]=[APC]+[CPD]+[APD]=\dfrac{1}{2}\cdot ab\cdot \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\cdot bc\cdot \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\cdot ca\cdot \dfrac{\sqrt{3}}{2}\cdot ca=(ab+bc+ca)\dfrac{\sqrt{3}}{4},$](http://latex.artofproblemsolving.com/f/3/0/f308ebac75ecf3097ff1b01b22ff5fc0926629b3.png)
so 
so 
so 
and 
and 
are cyclic. Note that ![\[ AX + BX + CX + DX + EX \le AX_{13} + CX_{13} + DX_{13} + BE \]](http://latex.artofproblemsolving.com/a/f/5/af5c183d3f9a2c63102c1e98c0544ed852b0208b.png)
by Fermat Point properties. Notably since 
opposite 
.
,
and 
and consider their intersection point 
,
is the format point. However, we also have 
,
!
,
and 
,
.
to get 
with equality holding when 
.
![\[CF = CX + \frac{AF \cdot DX + FD \cdot AX}{AD} = AX+CX+DX.\]](http://latex.artofproblemsolving.com/7/8/a/78af3c0d3d3e8188d2b2f96ef04836e822acd04a.png)
because 
and 
are cyclic further giving us ![\[\angle ACB = \angle AFB = \angle ADE = \angle AFE\]](http://latex.artofproblemsolving.com/b/8/a/b8ab0f95f7e0add9f292316930a4f07b8032dc1d.png)
,
and 
we get
![$[\triangle ACD]$](http://latex.artofproblemsolving.com/2/0/6/2066aabfd74e583d4df60454befd097a6386ab99.png)
in two ways we get 
using which we can get the value for 
![\[BE^2=AE^2+AB^2+2\cdot AE\cdot AB\cdot \frac{-11}{14}=1444\]](http://latex.artofproblemsolving.com/b/a/6/ba69f7a74dff0a58f0d502b01c8d40eda157dae1.png)
So 
and 
is our answer.