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Difference between revisions of "1990 IMO Problems/Problem 1"

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1. Chords AB and CD of a circle intersect at a point E inside the circle. Let M be an interior point of the segment EB. The tangent line at E to the circle
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==Problem==
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through D, E, and M intersects the lines BC and AC at F and G, respectively.
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Chords <math>AB</math> and <math>CD</math> of a circle intersect at a point <math>E</math> inside the circle. Let <math>M</math> be an interior point of the segment <math>\overline{EB}</math>. The tangent line at <math>E</math> to the circle through <math>D, E</math>, and <math>M</math> intersects the lines <math>\overline{BC}</math> and <math>{AC}</math> at <math>F</math> and <math>G</math>, respectively.
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If  <math>\frac{AM}{AB} = t</math>, find $\frac{EG}{EF} in terms of t.
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If  <math>\frac{AM}{AB} = t</math>, find <math>\frac{EG}{EF}</math> in terms of <math>t</math>.
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==Solution 1==
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With simple angle chasing, we find that triangles <math>CEG</math> and <math>BMD</math> are similar.
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so, <math>\frac{MB}{EC} = \frac{MD}{EG}</math>. .... <math>(1)</math>
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Again with simple angle chasing, we find that triangles <math>CEF</math> and <math>AMD</math> are similar.
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so, <math>\frac{MA}{EC} = \frac{MD}{EF}</math>. .... <math>(2)</math>
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so, by <math>(1)</math> and <math>(2)</math>, we have <math>\frac{EG}{EF} = \frac{MA}{MB} = \frac{t}{1-t}</math>.
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This solution was posted and copyrighted by e.lopes. The original thread for this problem can be found here: [https://aops.com/community/p366701]
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==Solution 2==
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This problem can be bashed with PoP and Ratio Lemma. Rewriting the given ratio gets <math>\frac{MA}{MB}=\frac{t}{1-t}</math>.
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By Ratio Lemma, <math>\frac{FB}{FC}=\frac{BE}{CE} \cdot \frac{\sin{\angle{FEB}}}{\sin{\angle{FEC}}}=\frac{DE}{AE} \cdot \frac{\sin{\angle{EDM}}}{\sin{\angle{DME}}}=\frac{DE}{AE} \cdot \frac{EM}{DE}=\frac{ME}{EA}</math>. Similarly, <math>\frac{GA}{GC}=\frac{ME}{EB}</math>. We can rewrite these equalities to get <math>\frac{AM}{EM}=\frac{BC}{FB}</math> and <math>\frac{BM}{EM}=\frac{AC}{GC}</math>.
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Using Ratio Lemma, <math>\frac{GE}{\sin{\angle{ACD}}}=\frac{GC}{\sin{\angle{GED}}}</math> and <math>\frac{EF}{\sin{\angle{BCD}}}=\frac{FC}{\sin{\angle{FEC}}}</math>. Since <math>\angle{GED}=\angle{FEC}</math>, we have <math>\frac{FE}{GE}=\frac{FC}{GC} \cdot \frac{\sin{\angle{BCD}}}{\sin{\angle{ACD}}}</math> (eq 1). Note that by Ratio Lemma, <math>\frac{\sin{\angle{BCD}}}{\sin{\angle{ACD}}}=\frac{CA}{CB} \cdot \frac{EB}{EA}</math>. Plugging this into (eq 1), we get <math>\frac{EF}{GE}=\frac{FC}{GC} \cdot \frac{CA}{CB} \cdot \frac{EB}{EA}=\frac{\frac{EA}{EM} \cdot FB}{\frac{EB}{EM} \cdot GA} \cdot \frac{CA}{CB} \cdot \frac{EB}{EA}=\frac{FB}{GA} \cdot \frac{CA}{CB}=\frac{EM}{AM} \cdot \frac{MB}{EM}=\frac{MB}{MA}=\frac{1-t}{t}</math>. So <math>\frac{EG}{EF}=\boxed{\frac{t}{1-t}}</math>.
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This solution was posted and copyrighted by AIME12345. The original thread for this problem can be found here: [https://aops.com/community/p11486878]
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== See Also == {{IMO box|year=1990|before=First Question|num-a=2}}

Latest revision as of 06:18, 17 November 2024

Problem

Chords $AB$ and $CD$ of a circle intersect at a point $E$ inside the circle. Let $M$ be an interior point of the segment $\overline{EB}$. The tangent line at $E$ to the circle through $D, E$, and $M$ intersects the lines $\overline{BC}$ and ${AC}$ at $F$ and $G$, respectively. If $\frac{AM}{AB} = t$, find $\frac{EG}{EF}$ in terms of $t$.

Solution 1

With simple angle chasing, we find that triangles $CEG$ and $BMD$ are similar.

so, $\frac{MB}{EC} = \frac{MD}{EG}$. .... $(1)$

Again with simple angle chasing, we find that triangles $CEF$ and $AMD$ are similar.

so, $\frac{MA}{EC} = \frac{MD}{EF}$. .... $(2)$

so, by $(1)$ and $(2)$, we have $\frac{EG}{EF} = \frac{MA}{MB} = \frac{t}{1-t}$.

This solution was posted and copyrighted by e.lopes. The original thread for this problem can be found here: [1]

Solution 2

This problem can be bashed with PoP and Ratio Lemma. Rewriting the given ratio gets $\frac{MA}{MB}=\frac{t}{1-t}$. By Ratio Lemma, $\frac{FB}{FC}=\frac{BE}{CE} \cdot \frac{\sin{\angle{FEB}}}{\sin{\angle{FEC}}}=\frac{DE}{AE} \cdot \frac{\sin{\angle{EDM}}}{\sin{\angle{DME}}}=\frac{DE}{AE} \cdot \frac{EM}{DE}=\frac{ME}{EA}$. Similarly, $\frac{GA}{GC}=\frac{ME}{EB}$. We can rewrite these equalities to get $\frac{AM}{EM}=\frac{BC}{FB}$ and $\frac{BM}{EM}=\frac{AC}{GC}$. Using Ratio Lemma, $\frac{GE}{\sin{\angle{ACD}}}=\frac{GC}{\sin{\angle{GED}}}$ and $\frac{EF}{\sin{\angle{BCD}}}=\frac{FC}{\sin{\angle{FEC}}}$. Since $\angle{GED}=\angle{FEC}$, we have $\frac{FE}{GE}=\frac{FC}{GC} \cdot \frac{\sin{\angle{BCD}}}{\sin{\angle{ACD}}}$ (eq 1). Note that by Ratio Lemma, $\frac{\sin{\angle{BCD}}}{\sin{\angle{ACD}}}=\frac{CA}{CB} \cdot \frac{EB}{EA}$. Plugging this into (eq 1), we get $\frac{EF}{GE}=\frac{FC}{GC} \cdot \frac{CA}{CB} \cdot \frac{EB}{EA}=\frac{\frac{EA}{EM} \cdot FB}{\frac{EB}{EM} \cdot GA} \cdot \frac{CA}{CB} \cdot \frac{EB}{EA}=\frac{FB}{GA} \cdot \frac{CA}{CB}=\frac{EM}{AM} \cdot \frac{MB}{EM}=\frac{MB}{MA}=\frac{1-t}{t}$. So $\frac{EG}{EF}=\boxed{\frac{t}{1-t}}$.

This solution was posted and copyrighted by AIME12345. The original thread for this problem can be found here: [2]

See Also

1990 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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