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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 14"

(Created page with "==Problem== Let there be a <math>\triangle ACD</math> such that <math>AC=5</math>, <math>AD=12</math>, and <math>CD=13</math>, and let <math>B</math> be a point on <math>AD</m...")
 
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To find <math>EC</math>, we can proceed by Power of a Point using point <math>D</math> on circle <math>(ABC)</math> to get <math>DE \cdot DC = DB \cdot DA.</math> Since <math>DC=13</math>, <math>DB = 7</math>, and <math>AD = 12</math>, we have <math>DE=\frac{84}{13}.</math> Since <math>CD=13</math>, we have <cmath>EC=CD-DE=\frac{85}{13} \qquad (2).</cmath>
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To find <math>EC</math>, we can proceed by Power of a Point using point <math>D</math> on circle <math>(ABC)</math> to get <math>DE \cdot DC = DB \cdot DA.</math> Since <math>DC=13</math>, <math>DB = 7</math>, and <math>AD = 12</math>, we have <math>DE=\frac{84}{13}.</math> Since <math>CD=13</math>, we have <cmath>EC=CD-DE=\frac{85}{13} \qquad (2).</cmath>  
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To find <math>BE</math>, we use the Pythagorean Theorem in <math>BED</math>. (We already found <math>\angle BEC=90^\circ</math>, which tells us that supplementary <math>\angle BED = 90^\circ</math>.) By the Pythagorean Theorem, <math>BE^2+DE^2=BD^2.</math> We found that <math>DE=\frac{84}{13}</math>, and since we are given <math>BD=7</math>, we have <cmath>BE=\sqrt{7^2-\left (\frac{84}{13}\right )^2}=\frac{35}{13} \qquad (3).</cmath>
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Our answer, by equation <math>(1)</math>, is <math>\frac{EC}{BE}</math>. From equation <math>(2),</math> <math>EC=\frac{85}{13}</math> and from equation <math>(3),</math> <math>BE=\frac{35}{13}</math>. Therefore, our final answer is <math>\frac{\frac{85}{13}}{\frac{35}{13}}=\frac{17}{7} \implies \boxed{24}.</math>
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~samrocksnature
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==Solution 2==
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Note <math>ABEC</math> is cyclic, so <math>\angle BEC = \angle BED = 90^o</math>. By Power Of A Point we have <math>7(7+5)=DE \cdot DC \implies DE=\frac{84}{13}</math>, <math>EC=\frac{85}{13}</math>. Now, note <cmath>\angle ABC = \angle BCA</cmath><cmath>\qquad =\angle BEA</cmath><cmath>= \angle AEC</cmath>Therefore, by Angle Bisector Theorem, <cmath>\frac{FC}{BF}=\frac{EC}{BE}=\frac{85}{35} \implies \frac{17}{7} \implies 17+7=\boxed{24}</cmath>
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~Geometry285
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==See also==
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#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
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#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 20:18, 11 July 2021

Problem

Let there be a $\triangle ACD$ such that $AC=5$, $AD=12$, and $CD=13$, and let $B$ be a point on $AD$ such that $BD=7.$ Let the circumcircle of $\triangle ABC$ intersect hypotenuse $CD$ at $E$ and $C$. Let $AE$ intersect $BC$ at $F$. If the ratio $\tfrac{FC}{BF}$ can be expressed as $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime, find $m+n.$

Solution

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$\mathbf{Lemma}:$ We claim that $EF$ is the angle bisector of $\angle BEC$.


$\mathbf{Proof:}$ Observe that $AB=5=AC$, which tells us that $\triangle ABC$ is a $45-45-90$ triangle. In cyclic quadrilateral $ABEC$, we have \[\angle ABC = 45^\circ = \angle AEC\] and \[\angle BAC+\angle BEC=180^\circ \implies \angle BEC = 90^\circ.\] Since $\angle BEA + \angle AEC = \angle BEC$, we have $\angle BEA =45^\circ = \angle AEC$. This means that $EA$, or equivalently $EF$, is an angle bisector of $\angle BEC$ in $\triangle BEC$.


$\mathbf{End~Proof}$


By the angle bisector theorem and our $\mathbf{Lemma},$ \[\frac{FC}{BF}=\frac{EC}{BE} \qquad (1).\] We seek the lengths $EC$ and $BE$.


To find $EC$, we can proceed by Power of a Point using point $D$ on circle $(ABC)$ to get $DE \cdot DC = DB \cdot DA.$ Since $DC=13$, $DB = 7$, and $AD = 12$, we have $DE=\frac{84}{13}.$ Since $CD=13$, we have \[EC=CD-DE=\frac{85}{13} \qquad (2).\]


To find $BE$, we use the Pythagorean Theorem in $BED$. (We already found $\angle BEC=90^\circ$, which tells us that supplementary $\angle BED = 90^\circ$.) By the Pythagorean Theorem, $BE^2+DE^2=BD^2.$ We found that $DE=\frac{84}{13}$, and since we are given $BD=7$, we have \[BE=\sqrt{7^2-\left (\frac{84}{13}\right )^2}=\frac{35}{13} \qquad (3).\]


Our answer, by equation $(1)$, is $\frac{EC}{BE}$. From equation $(2),$ $EC=\frac{85}{13}$ and from equation $(3),$ $BE=\frac{35}{13}$. Therefore, our final answer is $\frac{\frac{85}{13}}{\frac{35}{13}}=\frac{17}{7} \implies \boxed{24}.$

~samrocksnature

Solution 2

Note $ABEC$ is cyclic, so $\angle BEC = \angle BED = 90^o$. By Power Of A Point we have $7(7+5)=DE \cdot DC \implies DE=\frac{84}{13}$, $EC=\frac{85}{13}$. Now, note \[\angle ABC = \angle BCA\]\[\qquad =\angle BEA\]\[= \angle AEC\]Therefore, by Angle Bisector Theorem, \[\frac{FC}{BF}=\frac{EC}{BE}=\frac{85}{35} \implies \frac{17}{7} \implies 17+7=\boxed{24}\]

~Geometry285

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

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