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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 7"
 
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Line 12: Line 12:
 
Note that <math>PA = 3</math> and <math>PD = 3 + 8 = 11</math>, so
 
Note that <math>PA = 3</math> and <math>PD = 3 + 8 = 11</math>, so
 
<cmath>[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.</cmath>
 
<cmath>[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.</cmath>
−
 
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<asy>
 
−
size(4cm);
 
−
 
−
pair o = (0,0);
 
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pair p = (sqrt(58), 0);
 
−
 
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path out = circle(o, 5);
 
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path inn = circle(o, 3);
 
−
 
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filldraw(inn, lightgreen+opacity(0.2), green);
 
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filldraw(out, lightgreen+opacity(0.2), green);
 
−
 
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pair[] ac = intersectionpoints(circle(p, 3), out);
 
−
 
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pair a = ac[0];
 
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pair c = ac[1];
 
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pair b = intersectionpoints(p--a, out)[0];
 
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pair d = intersectionpoints(p--c, out)[1];
 
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pair m = intersectionpoints(p--a, inn)[0];
 
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pair n = intersectionpoints(p--c, inn)[0];
 
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pair t = intersectionpoint(o--p, a--c);
 
−
 
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draw(b--p--d, blue);
 
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draw(n--o--m, dashed+green);
 
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draw(c--a, blue+dashed);
 
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draw(o--p, blue+dashed);
 
−
 
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dot("$A$", a, dir(345));
 
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dot("$C$", c, dir(55));
 
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dot("$B$", b, dir(240));
 
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dot("$D$", d, dir(135));
 
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dot("$M$", m, dir(330));
 
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dot("$N$", n, dir(60));
 
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dot("$T$", t, dir(135));
 
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dot("$P$", p, dir(45));
 
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dot("$O$", o, dir(45));
 
−
 
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clip((20,20)--(-20,20)--(-20,-20)--(20,-20)--cycle);
 
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</asy>
 
  
 
We now proceed with the other case.
 
We now proceed with the other case.
Line 64: Line 24:
 
<cmath>[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.</cmath>
 
<cmath>[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.</cmath>
  
−
<asy>
+
The sum of the areas is thus <math>\frac{477}{29}</math> and the answer is <math>\boxed{506}</math>.
−
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];
 
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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];
 
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pair t = intersectionpoint(o--p, a--c);
 
−
 
 
−
draw(a--b, blue);
 
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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));
 
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dot("$M$", m, dir(135));
 
−
dot("$N$", n, dir(225));
 
−
dot("$T$", t, dir(0));
 
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dot("$P$", p, dir(45));
 
−
dot("$O$", o, dir(45));
 
−
 
 
−
clip((20,20)--(-20,20)--(-20,-20)--(20,-20)--cycle);
 
−
</asy>
 
  
−
The sum of the areas is thus <math>\frac{477}{29}</math> and the answer is <math>\boxed{506}</math>
+
~SMO_Team

Latest revision as of 21:10, 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 \[[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.\]

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 \[[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.\]

The sum of the areas is thus $\frac{477}{29}$ and the answer is $\boxed{506}$.

~SMO_Team

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