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

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With simple angle chasing, we find that triangles <math>CEG</math> and <math>BMD</math> are similar.
 
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>. (*)
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so, <math>\frac{MB}{EC} = \frac{MD}{EG}</math>. .... <math>(1)</math>
  
 
Again with simple angle chasing, we find that triangles <math>CEF</math> and <math>AMD</math> are similar.
 
Again with simple angle chasing, we find that triangles <math>CEF</math> and <math>AMD</math> are similar.
  
−
so, <math>/frac{MA}{EC} = /frac{MD}{EF}</math>. (**)
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so, <math>\frac{MA}{EC} = \frac{MD}{EF}</math>. .... <math>(2)</math>
  
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so, by (*) and (**), we have <math>/frac{EG}{EF} = /frac{MA}{MB} = /frac{t}{1-t}</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>.
  
 
This solution was posted and copyrighted by e.lopes. The original thread for this problem can be found here: [https://aops.com/community/p366701]
 
This solution was posted and copyrighted by e.lopes. The original thread for this problem can be found here: [https://aops.com/community/p366701]

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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