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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 7"
Line 15: Line 15:
 
<asy>
 
<asy>
 
size(4cm);
 
size(4cm);
−
point o, p, a, c, b, d, m, n, t;
 
−
o = (0,0);
 
−
p = (sqrt(58), 0);
 
  
−
circle out, in;
+
pair o = (0,0);
−
out = circle(o, 5);
+
pair p = (sqrt(58), 0);
−
in = circle(o, 3);
 
  
−
filldraw(in, opacity(0.2)+lightgreen,green);
+
path out = circle(o, 5);
−
filldraw(out, opacity(0.2)+lightgreen,green);
+
path inn = circle(o, 3);
  
−
point[] ac = intersectionpoints(circle(p, 3), out);
+
filldraw(inn, lightgreen+opacity(0.2), green);
 +
filldraw(out, lightgreen+opacity(0.2), green);
  
−
a = ac[0];
+
pair[] ac = intersectionpoints(circle(p, 3), out);
−
c = ac[1];
+
 
−
b = intersectionpoints(line(p,a), out)[0];
+
pair a = ac[0];
−
d = intersectionpoints(line(p,c), out)[1];
+
pair c = ac[1];
−
m = intersectionpoints(line(p,a), in)[0];
+
pair b = intersectionpoints(p--a, out)[0];
−
n = intersectionpoints(line(p,c), in)[0];
+
pair d = intersectionpoints(p--c, out)[1];
−
t = intersectionpoint(line(o,p), line(a,c));
+
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(b--p--d, blue);
Line 45: Line 44:
 
dot("$B$", b, dir(240));
 
dot("$B$", b, dir(240));
 
dot("$D$", d, dir(135));
 
dot("$D$", d, dir(135));
−
 
 
dot("$M$", m, dir(330));
 
dot("$M$", m, dir(330));
 
dot("$N$", n, dir(60));
 
dot("$N$", n, dir(60));
 
dot("$T$", t, dir(135));
 
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);
 +
</asy>
 +
 +
We now proceed with the other case.
 +
 +
We solve to get <math>OP = \sqrt{10}</math> and <math>PM = PN = 1</math>.
 +
 +
Once again, <math>\triangle NOP \sim \triangle CTP</math>, and since
 +
<cmath>[NOP] = \frac{1}{2} \cdot 3 \cdot 1 = \frac{3}{2},</cmath>
 +
it follows that
 +
<cmath>[CTP] = [NOP] \cdot \left(\frac{CP}{OP}\right)^2 = \frac{3}{2} \cdot \frac{9}{10} = \frac{27}{20},</cmath>
 +
so
 +
<cmath>[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.</cmath>
 +
 +
<asy>
 +
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("$P$", p, dir(45));
 
dot("$O$", o, dir(45));
 
dot("$O$", o, dir(45));
Line 55: Line 104:
 
clip((20,20)--(-20,20)--(-20,-20)--(20,-20)--cycle);
 
clip((20,20)--(-20,20)--(-20,-20)--(20,-20)--cycle);
 
</asy>
 
</asy>
 +
 +
The sum of the areas is thus <math>\frac{477}{29}</math> and the answer is <math>\boxed{506}</math>

Revision as of 21:07, 9 September 2025

Problem

Concentric circles $\omega$ and $\omega_1$ are drawn, with radii $3$ and $5,$ respectively. Chords $AB$ and $CD$ of $\omega_1$ are both tangent to $\omega$ and intersect at $P.$ If $PA=PC = 3,$ then the sum of all possible distinct values of $[PAD]$ can be expressed as $\frac{m}{n},$ for relatively prime positive integers $m$ and $n.$ Find $m+n.$

Solution

There are two cases. First, consider the case where $P$ lies outside the circle.

Using the Pythagorean theorem, we find that $AB = CD = 2\sqrt{5^2 - 3^2} = 8$. We also have $OP = \sqrt{3^2 + 7^2} = \sqrt{58}$.

Let $T$ be the midpoint of $\overline{CA}$, which lies on $\overline{PA}$ by symmetry. Then $\triangle NOP \sim \triangle CTP$, so since \[[NOP] = \frac{1}{2} \cdot 3 \cdot 7 = \frac{21}{2},\] it follows that \[[CTP] = [NOP] \cdot \left(\frac{CP}{OP}\right)^2 = \frac{21}{2} \cdot \frac{9}{58} = \frac{189}{116},\] and thus \[[PAC] = 2 \cdot [CTP] = \frac{189}{58}.\]

Note that $PA = 3$ and $PD = 3 + 8 = 11$, so \[[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.\] 

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 $OP = \sqrt{10}$ and $PM = PN = 1$.

Once again, $\triangle NOP \sim \triangle CTP$, and since \[[NOP] = \frac{1}{2} \cdot 3 \cdot 1 = \frac{3}{2},\] it follows that \[[CTP] = [NOP] \cdot \left(\frac{CP}{OP}\right)^2 = \frac{3}{2} \cdot \frac{9}{10} = \frac{27}{20},\] so \[[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.\] 

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 $\frac{477}{29}$ and the answer is $\boxed{506}$

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