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Difference between revisions of "2021 WSMO Speed Round Problems/Problem 7"

(Created page with "==Problem== Consider triangle <math>ABC</math> with side lengths <math>AB=13,AC=14,BC=15</math> and incircle <math>\omega</math>. A second circle <math>\omega_2</math> is draw...")
 
(→‎Solution)
 
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==Solution==
 
==Solution==
 
Note that the length of the <math>A</math>-angle bisector is <math>\frac{\sqrt{(b+c-a)(b+c+a)bc}}{b+c}.</math> Now, let <math>I</math> be the incenter of triangle <math>ABC.</math> This means that <cmath>AI=\frac{b+c}{a+b+c}\cdot\frac{\sqrt{b+c-a}(b+c+a)bc}{b+c}=\frac{\sqrt{(b+c-a)(b+c+a)bc}}{a+b+c}=\frac{\sqrt{12\cdot42\cdot13\cdot14}}{42}=2\sqrt{13}.</cmath> Now, from Heron's formula, we find that the area of triangle <math>ABC</math> is <cmath>\sqrt{\left(\frac{\left(13+14+15\right)}{2}\right)\left(\frac{\left(13+14+15\right)}{2}-13\right)\left(\frac{\left(13+14+15\right)}{2}-14\right)\left(\frac{\left(13+14+15\right)}{2}-15\right)}=\sqrt{21\cdot6\cdot7\cdot8}=84.</cmath> Since the area of a triangle is the product of the semi-perimeter and the inradius, we find that the length of the inradius of triangle <math>ABC</math> is <math>\frac{84}{\frac{13+14+15}{2}}=\frac{84}{21}=4.</math> Now, let the radius of <math>\omega_2</math> be <math>r.</math> From similar triangles, we find that <cmath>\frac{r}{4}=\frac{2\sqrt{13}-4-r}{2\sqrt{13}}\Leftrightarrow r=\frac{4(2\sqrt{13}-4)}{2\sqrt{13}+4}=\frac{4(2\sqrt{13}-4)^2}{(2\sqrt{13}-4)(2\sqrt{13}+4)}=\frac{4(68-16\sqrt13)}{(2\sqrt{13})^2-4^2}=\frac{4(68-16\sqrt{13})}{36}=\frac{68-16\sqrt{13}}{9}\Longrightarrow68+16+13+9=\boxed{106}.</cmath>
 
Note that the length of the <math>A</math>-angle bisector is <math>\frac{\sqrt{(b+c-a)(b+c+a)bc}}{b+c}.</math> Now, let <math>I</math> be the incenter of triangle <math>ABC.</math> This means that <cmath>AI=\frac{b+c}{a+b+c}\cdot\frac{\sqrt{b+c-a}(b+c+a)bc}{b+c}=\frac{\sqrt{(b+c-a)(b+c+a)bc}}{a+b+c}=\frac{\sqrt{12\cdot42\cdot13\cdot14}}{42}=2\sqrt{13}.</cmath> Now, from Heron's formula, we find that the area of triangle <math>ABC</math> is <cmath>\sqrt{\left(\frac{\left(13+14+15\right)}{2}\right)\left(\frac{\left(13+14+15\right)}{2}-13\right)\left(\frac{\left(13+14+15\right)}{2}-14\right)\left(\frac{\left(13+14+15\right)}{2}-15\right)}=\sqrt{21\cdot6\cdot7\cdot8}=84.</cmath> Since the area of a triangle is the product of the semi-perimeter and the inradius, we find that the length of the inradius of triangle <math>ABC</math> is <math>\frac{84}{\frac{13+14+15}{2}}=\frac{84}{21}=4.</math> Now, let the radius of <math>\omega_2</math> be <math>r.</math> From similar triangles, we find that <cmath>\frac{r}{4}=\frac{2\sqrt{13}-4-r}{2\sqrt{13}}\Leftrightarrow r=\frac{4(2\sqrt{13}-4)}{2\sqrt{13}+4}=\frac{4(2\sqrt{13}-4)^2}{(2\sqrt{13}-4)(2\sqrt{13}+4)}=\frac{4(68-16\sqrt13)}{(2\sqrt{13})^2-4^2}=\frac{4(68-16\sqrt{13})}{36}=\frac{68-16\sqrt{13}}{9}\Longrightarrow68+16+13+9=\boxed{106}.</cmath>
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~pinkpig

Latest revision as of 10:59, 23 December 2021

Problem

Consider triangle $ABC$ with side lengths $AB=13,AC=14,BC=15$ and incircle $\omega$. A second circle $\omega_2$ is drawn which is tangent to $AB,AC$ and externally tangent to $\omega$. The radius of $\omega_2$ can be expressed as $\frac{a-b\sqrt{c}}{d}$, where $\gcd{(a,b,d)}=1$ and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.

Solution

Note that the length of the $A$-angle bisector is $\frac{\sqrt{(b+c-a)(b+c+a)bc}}{b+c}.$ Now, let $I$ be the incenter of triangle $ABC.$ This means that \[AI=\frac{b+c}{a+b+c}\cdot\frac{\sqrt{b+c-a}(b+c+a)bc}{b+c}=\frac{\sqrt{(b+c-a)(b+c+a)bc}}{a+b+c}=\frac{\sqrt{12\cdot42\cdot13\cdot14}}{42}=2\sqrt{13}.\] Now, from Heron's formula, we find that the area of triangle $ABC$ is \[\sqrt{\left(\frac{\left(13+14+15\right)}{2}\right)\left(\frac{\left(13+14+15\right)}{2}-13\right)\left(\frac{\left(13+14+15\right)}{2}-14\right)\left(\frac{\left(13+14+15\right)}{2}-15\right)}=\sqrt{21\cdot6\cdot7\cdot8}=84.\] Since the area of a triangle is the product of the semi-perimeter and the inradius, we find that the length of the inradius of triangle $ABC$ is $\frac{84}{\frac{13+14+15}{2}}=\frac{84}{21}=4.$ Now, let the radius of $\omega_2$ be $r.$ From similar triangles, we find that \[\frac{r}{4}=\frac{2\sqrt{13}-4-r}{2\sqrt{13}}\Leftrightarrow r=\frac{4(2\sqrt{13}-4)}{2\sqrt{13}+4}=\frac{4(2\sqrt{13}-4)^2}{(2\sqrt{13}-4)(2\sqrt{13}+4)}=\frac{4(68-16\sqrt13)}{(2\sqrt{13})^2-4^2}=\frac{4(68-16\sqrt{13})}{36}=\frac{68-16\sqrt{13}}{9}\Longrightarrow68+16+13+9=\boxed{106}.\]

~pinkpig

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