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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 7"
Line 16: Line 16:
 
size(4cm);
 
size(4cm);
  
−
pair o = (0,0);
+
pair O = (0,0);
−
pair p = (sqrt(58), 0);
+
pair P = (7.6158,0);
  
−
path out = circle(o, 5);
+
filldraw(circle(O,3), lightgreen, green);
−
path inn = circle(o, 3);
+
filldraw(circle(O,5), lightgreen, green);
  
−
filldraw(inn, lightgreen+opacity(0.2), green);
+
pair A = (6.9, 2.6);
−
filldraw(out, lightgreen+opacity(0.2), green);
+
pair C = (6.9, -2.6);
 +
pair B = (9.7, 3.8);
 +
pair D = (9.7, -3.8);
 +
pair M = (4.1, 1.6);
 +
pair N = (4.1, -1.6);
 +
pair T = (3.5, 0);
  
−
pair[] ac = intersectionpoints(circle(p, 3), out);
+
draw(B--P--D, blue);
 +
draw(N--O--M, dashed+green);
 +
draw(C--A, dashed+blue);
 +
draw(O--P, dashed+blue);
  
−
pair a = ac[0];
+
dot("$A$", A, dir(345));
−
pair c = ac[1];
+
dot("$C$", C, dir(55));
−
pair b = intersectionpoints(p--a, out)[0];
+
dot("$B$", B, dir(240));
−
pair d = intersectionpoints(p--c, out)[1];
+
dot("$D$", D, dir(135));
−
pair m = intersectionpoints(p--a, inn)[0];
+
dot("$M$", M, dir(330));
−
pair n = intersectionpoints(p--c, inn)[0];
+
dot("$N$", N, dir(60));
−
pair t = intersectionpoint(o--p, a--c);
+
dot("$T$", T, dir(135));
−
 
+
dot("$P$", P, dir(45));
−
draw(b--p--d, blue);
+
dot("$O$", O, dir(45));
−
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);
 
clip((20,20)--(-20,20)--(-20,-20)--(20,-20)--cycle);
Line 67: Line 62:
 
size(4cm);
 
size(4cm);
  
−
pair o = (0,0);
+
pair O = (0,0);
−
pair p = (sqrt(10), 0);
+
pair P = (3.16, 0);  
  
−
path out = circle(o, 5);
+
path outer = Circle(O, 5);
−
path inn = circle(o, 3);
+
path inner = Circle(O, 3);
  
−
filldraw(inn, lightgreen + opacity(0.2), green);
+
filldraw(outer, lightgreen, green);
−
filldraw(out, lightgreen + opacity(0.2), green);
+
filldraw(inner, lightgreen, green);
  
−
pair[] ac = intersectionpoints(circle(p, 3), out);
+
pair A = (2.2, 2.6);
 +
pair B = (5, 6);
 +
pair C = (2.2, -2.6);
 +
pair D = (5, -6);
 +
pair M = (1.3, 1.6);
 +
pair N = (1.3, -1.6);
 +
pair T = (1.1, 0);  
  
−
pair a = ac[0];
+
draw(A--B, blue);
−
pair c = ac[1];
+
draw(C--D, blue);
−
pair b = intersectionpoints(p--a, out)[1];
+
draw(N--O--M, dashed+green);
−
pair d = intersectionpoints(p--c, out)[0];
+
draw(A--C, blue+dashed);
−
pair m = intersectionpoints(p--a, inn)[0];
+
draw(O--P, blue+dashed);
−
pair n = intersectionpoints(p--c, inn)[0];
 
−
pair t = intersectionpoint(o--p, a--c);
 
  
−
draw(a--b, blue);
+
dot("$A$", A, NE);
−
draw(c--d, blue);
+
dot("$B$", B, NE);
−
draw(n--o--m, dashed + green);
+
dot("$C$", C, SE);
−
draw(c--a, blue + dashed);
+
dot("$D$", D, SE);
−
draw(o--p, blue + dashed);
+
dot("$M$", M, N);
−
 
+
dot("$N$", N, S);
−
dot("$A$", a, dir(345));
+
dot("$T$", T, dir(0));
−
dot("$C$", c, dir(55));
+
dot("$O$", O, dir(135));
−
dot("$B$", b, dir(240));
+
dot("$P$", P, dir(45));
−
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);
 
 
</asy>
 
</asy>
  
 
The sum of the areas is thus <math>\frac{477}{29}</math> and the answer is <math>\boxed{506}</math>
 
The sum of the areas is thus <math>\frac{477}{29}</math> and the answer is <math>\boxed{506}</math>

Revision as of 21:09, 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 \[[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.\]

[asy] size(4cm); pair O = (0,0); pair P = (7.6158,0); filldraw(circle(O,3), lightgreen, green); filldraw(circle(O,5), lightgreen, green); pair A = (6.9, 2.6); pair C = (6.9, -2.6); pair B = (9.7, 3.8); pair D = (9.7, -3.8); pair M = (4.1, 1.6); pair N = (4.1, -1.6); pair T = (3.5, 0); draw(B--P--D, blue); draw(N--O--M, dashed+green); draw(C--A, dashed+blue); draw(O--P, dashed+blue); 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); [/asy]

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

[asy] size(4cm); pair O = (0,0); pair P = (3.16, 0); path outer = Circle(O, 5); path inner = Circle(O, 3); filldraw(outer, lightgreen, green); filldraw(inner, lightgreen, green); pair A = (2.2, 2.6); pair B = (5, 6); pair C = (2.2, -2.6); pair D = (5, -6); pair M = (1.3, 1.6); pair N = (1.3, -1.6); pair T = (1.1, 0); draw(A--B, blue); draw(C--D, blue); draw(N--O--M, dashed+green); draw(A--C, blue+dashed); draw(O--P, blue+dashed); dot("$A$", A, NE); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, SE); dot("$M$", M, N); dot("$N$", N, S); dot("$T$", T, dir(0)); dot("$O$", O, dir(135)); dot("$P$", P, dir(45)); [/asy]

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

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