2023 SSMO Accuracy Round Problems/Problem 7
Problem
Concentric circles 
and 
are drawn, with radii 
and 
respectively. Chords 
and 
of are both tangent to and intersect at 
If 
then the sum of all possible distinct values of ![$[PAD]$](http://latex.artofproblemsolving.com/a/7/e/a7e21efb4fd61ce8b01083b7ab8562bccfcb8406.png)
can be expressed as 
for relatively prime positive integers 
and 
Find 
Solution
There are two cases. First, consider the case where 
lies outside the circle.
Using the Pythagorean theorem, we find that 
. We also have 
.
Let 
be the midpoint of 
, which lies on 
by symmetry. Then 
, so since ![\[[NOP] = \frac{1}{2} \cdot 3 \cdot 7 = \frac{21}{2},\]](http://latex.artofproblemsolving.com/b/2/d/b2d9803929df1b36918db625e2c3501ef4c5e255.png)
it follows that
![\[[CTP] = [NOP] \cdot \left(\frac{CP}{OP}\right)^2 = \frac{21}{2} \cdot \frac{9}{58} = \frac{189}{116},\]](http://latex.artofproblemsolving.com/b/3/6/b365a9cb01a7a18244c1cd2fe31134764b6cb240.png)
and thus ![\[[PAC] = 2 \cdot [CTP] = \frac{189}{58}.\]](http://latex.artofproblemsolving.com/f/4/9/f49266288f37838d5bb6469196b5ebe129938993.png)
Note that 
and 
, so
![\[[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.\]](http://latex.artofproblemsolving.com/c/6/a/c6a8ab5b68533ebb1a674599f800414118e5ce64.png)
size(4cm);
pair o = (0,0);
pair p = (sqrt(58), 0);
path out = circle(o, 5);
path inn = circle(o, 3);
filldraw(inn, lightgreen+opacity(0.2), green);
filldraw(out, lightgreen+opacity(0.2), green);
pair[] ac = intersectionpoints(circle(p, 3), out);
pair a = ac[0];
pair c = ac[1];
pair b = intersectionpoints(p--a, out)[0];
pair d = intersectionpoints(p--c, out)[1];
pair m = intersectionpoints(p--a, inn)[0];
pair n = intersectionpoints(p--c, inn)[0];
pair t = intersectionpoint(o--p, a--c);
draw(b--p--d, blue);
draw(n--o--m, dashed+green);
draw(c--a, blue+dashed);
draw(o--p, blue+dashed);
dot("$A$", a, dir(345));
dot("$C$", c, dir(55));
dot("$B$", b, dir(240));
dot("$D$", d, dir(135));
dot("$M$", m, dir(330));
dot("$N$", n, dir(60));
dot("$T$", t, dir(135));
dot("$P$", p, dir(45));
dot("$O$", o, dir(45));
clip((20,20)--(-20,20)--(-20,-20)--(20,-20)--cycle);
(Error making remote request. Unknown error_msg)
We now proceed with the other case.
We solve to get 
and 
.
Once again, , and since
![\[[NOP] = \frac{1}{2} \cdot 3 \cdot 1 = \frac{3}{2},\]](http://latex.artofproblemsolving.com/7/4/5/74547544930936d41d83d4196d38f3b2e4ab43a7.png)
it follows that
![\[[CTP] = [NOP] \cdot \left(\frac{CP}{OP}\right)^2 = \frac{3}{2} \cdot \frac{9}{10} = \frac{27}{20},\]](http://latex.artofproblemsolving.com/c/8/2/c82b5b746d29c7cb6813a36d81d01831d251a5d4.png)
so
![\[[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.\]](http://latex.artofproblemsolving.com/1/e/7/1e74477c3729a696cb614dfebd759be18b805956.png)
size(4cm);
pair o = (0,0);
pair p = (sqrt(10), 0);
path out = circle(o, 5);
path inn = circle(o, 3);
filldraw(inn, lightgreen + opacity(0.2), green);
filldraw(out, lightgreen + opacity(0.2), green);
pair[] ac = intersectionpoints(circle(p, 3), out);
pair a = ac[0];
pair c = ac[1];
pair b = intersectionpoints(p--a, out)[1];
pair d = intersectionpoints(p--c, out)[0];
pair m = intersectionpoints(p--a, inn)[0];
pair n = intersectionpoints(p--c, inn)[0];
pair t = intersectionpoint(o--p, a--c);
draw(a--b, blue);
draw(c--d, blue);
draw(n--o--m, dashed + green);
draw(c--a, blue + dashed);
draw(o--p, blue + dashed);
dot("$A$", a, dir(345));
dot("$C$", c, dir(55));
dot("$B$", b, dir(240));
dot("$D$", d, dir(135));
dot("$M$", m, dir(135));
dot("$N$", n, dir(225));
dot("$T$", t, dir(0));
dot("$P$", p, dir(45));
dot("$O$", o, dir(45));
clip((20,20)--(-20,20)--(-20,-20)--(20,-20)--cycle);
(Error making remote request. Unknown error_msg)
The sum of the areas is thus 
and the answer is 
