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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 7"

(Created page with "==Problem== Concentric circles <math>\omega</math> and <math>\omega_1</math> are drawn, with radii <math>3</math> and <math>5,</math> respectively. Chords <math>AB</math> and...")
 
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
 +
There are two cases. First, consider the case where <math>P</math> lies outside the circle.
 +
 +
Using the Pythagorean theorem, we find that <math>AB = CD = 2\sqrt{5^2 - 3^2} = 8</math>. We also have <math>OP = \sqrt{3^2 + 7^2} = \sqrt{58}</math>.
 +
 +
Let <math>T</math> be the midpoint of <math>\overline{CA}</math>, which lies on <math>\overline{PA}</math> by symmetry. Then <math>\triangle NOP \sim \triangle CTP</math>, so since <cmath>[NOP] = \frac{1}{2} \cdot 3 \cdot 7 = \frac{21}{2},</cmath> it follows that
 +
<cmath>[CTP] = [NOP] \cdot \left(\frac{CP}{OP}\right)^2 = \frac{21}{2} \cdot \frac{9}{58} = \frac{189}{116},</cmath> and thus <cmath>[PAC] = 2 \cdot [CTP] = \frac{189}{58}.</cmath>
 +
 +
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>
 +
 +
<asy>
 +
size(4cm);
 +
point o, p, a, c, b, d, m, n, t;
 +
o = (0,0);
 +
p = (sqrt(58), 0);
 +
 +
circle out, in;
 +
out = circle(o, 5);
 +
in = circle(o, 3);
 +
 +
filldraw(in, opacity(0.2)+lightgreen,green);
 +
filldraw(out, opacity(0.2)+lightgreen,green);
 +
 +
point[] ac = intersectionpoints(circle(p, 3), out);
 +
 +
a = ac[0];
 +
c = ac[1];
 +
b = intersectionpoints(line(p,a), out)[0];
 +
d = intersectionpoints(line(p,c), out)[1];
 +
m = intersectionpoints(line(p,a), in)[0];
 +
n = intersectionpoints(line(p,c), in)[0];
 +
t = intersectionpoint(line(o,p), line(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);
 +
</asy>

Revision as of 21:06, 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);
point o, p, a, c, b, d, m, n, t;
o = (0,0);
p = (sqrt(58), 0);

circle out, in;
out = circle(o, 5);
in = circle(o, 3);

filldraw(in, opacity(0.2)+lightgreen,green);
filldraw(out, opacity(0.2)+lightgreen,green);

point[] ac = intersectionpoints(circle(p, 3), out);

a = ac[0];
c = ac[1];
b = intersectionpoints(line(p,a), out)[0];
d = intersectionpoints(line(p,c), out)[1];
m = intersectionpoints(line(p,a), in)[0];
n = intersectionpoints(line(p,c), in)[0];
t = intersectionpoint(line(o,p), line(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);
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