Difference between revisions of "1990 IMO Problems/Problem 1"
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− | + | ==Problem== | |
− | through D, E, and M intersects the lines BC and AC at F and G, respectively. | + | 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. |
− | If <math>\frac{AM}{AB} = t</math>, find | + | If <math>\frac{AM}{AB} = t</math>, find <math>\frac{EG}{EF}</math> in terms of <math>t</math>. |
+ | |||
+ | ==Solution 1== | ||
+ | With simple angle chasing, we find that triangles <math>CEG</math> and <math>MDB</math> are similar. | ||
+ | |||
+ | so, <math>GE/EC = MD/MB</math>. (*) | ||
+ | |||
+ | again with simple angle chasing, we find that triangles <math>CEF</math> and <math>AMD</math> are similar. | ||
+ | |||
+ | so, <math>EF/DM = CE/AM</math>. (**) | ||
+ | |||
+ | so, by (*) and (**), we have <math>GE/EF = MA/MB = t/(t-1)</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] | ||
+ | |||
+ | ==Solution 2== | ||
+ | This problem can be bashed with PoP and Ratio Lemma. Rewriting the given ratio gets <math>\frac{MA}{MB}=\frac{t}{1-t}</math>. | ||
+ | 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>. | ||
+ | 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>. | ||
+ | |||
+ | This solution was posted and copyrighted by AIME12345. The original thread for this problem can be found here: [https://aops.com/community/p11486878] | ||
+ | |||
+ | == See Also == {{IMO box|year=1990|before=First Question|num-a=2}} |
Latest revision as of 13:34, 30 January 2021
Contents
Problem
Chords and
of a circle intersect at a point
inside the circle. Let
be an interior point of the segment
. The tangent line at
to the circle through
, and
intersects the lines
and
at
and
, respectively.
If
, find
in terms of
.
Solution 1
With simple angle chasing, we find that triangles and
are similar.
so, . (*)
again with simple angle chasing, we find that triangles and
are similar.
so, . (**)
so, by (*) and (**), we have .
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 .
By Ratio Lemma,
. Similarly,
. We can rewrite these equalities to get
and
.
Using Ratio Lemma,
and
. Since
, we have
(eq 1). Note that by Ratio Lemma,
. Plugging this into (eq 1), we get
. So
.
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 |