Difference between revisions of "2023 USAJMO Problems/Problem 6"
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Latest revision as of 12:41, 2 June 2024
Problem
Isosceles triangle 
, with 
, is inscribed in circle 
. Let 
be an arbitrary point inside 
such that 
. Ray 
intersects again at 
(other than 
). Point 
(other than ) is chosen on such that 
. Line 
intersects rays 
and 
at points 
and 
, respectively. Prove that 
.
Solution
All angle and side length names are defined as in the figures below. Figure 1 is the diagram of the problem while Figure 2 is the diagram of the Ratio Lemma. Do note that the point names defined in the Ratio Lemma are not necessarily the same defined points on Figure 1.
First, we claim the Ratio Lemma:
![\[\frac{\sin{\theta_2}}{\sin{\theta_2’}}=\frac{PB}{PC}\cdot\frac{b}{c}.\]](http://latex.artofproblemsolving.com/f/d/8/fd8b8f24e059cd7deb80e41cc0d7a564291f6a54.png)
We prove this as follows:
By the Law of Sines of 
we get:
![\[\frac{\sin\theta_2}{BP}=\frac{\sin\theta_1}{c}=\frac{\sin B}{AP}.\]](http://latex.artofproblemsolving.com/6/5/6/6562a868a03e93aed8ec800b8c690083249d37b9.png)
By the Law of Sines of 
we get:
![\[\frac{\sin\theta_2’}{PC}=\frac{\sin\theta_1’}{b}=\frac{\sin C}{AP}.\]](http://latex.artofproblemsolving.com/5/1/4/51400526e75694a6110530ad44f700b1c4c3592b.png)
Dividing each of the corresponding equations, we get:
![\[\frac{PC}{PB}\cdot\frac{\sin\theta_2}{\sin\theta_1}=\frac{b}{c}\cdot\frac{\sin\theta_1}{\sin\theta_1’}=\frac{\sin B}{\sin C}.\]](http://latex.artofproblemsolving.com/e/d/b/edb5f148a7915ffb1624f22c85bf06dff753775c.png)
Noting that 
we have:
![\[\frac{\sin\theta_2}{\sin\theta_1}=\frac{PB}{PC}\cdot\frac{b}{c}.\]](http://latex.artofproblemsolving.com/a/a/4/aa429c8f441e586c77e71ae269e4ef0527ca0ca8.png)
We transition into our problem.
By Power of a Point on circle 
we have:
![\[XK\cdot XD=XF\cdot XE\]](http://latex.artofproblemsolving.com/9/f/a/9fa54c74e9b4d2724651f2ced184912d55506089.png)
![\[YL\cdot YD=YF\cdot YE.\]](http://latex.artofproblemsolving.com/5/b/2/5b27a4cbc4f8fa552d63c4ac5f5f929c85b282a3.png)
By power of a point on circle 
we have:
![\[XB\cdot XA=XF\cdot XE\]](http://latex.artofproblemsolving.com/8/7/d/87d0157572921b1e087a28f693d6c104016e584c.png)
![\[YC\cdot YA=YF\cdot YE.\]](http://latex.artofproblemsolving.com/9/c/2/9c2485400a3888ff47be29486e86bb8fff9f5eb0.png)
Using the transitive property, we have:
![\[XK\cdot XD=XB\cdot XA\]](http://latex.artofproblemsolving.com/d/e/3/de3ecccc4653aa561f9818d02a34bdc1acfdb11a.png)
![\[YC\cdot YA=YL\cdot YD.\]](http://latex.artofproblemsolving.com/4/4/8/448492dd0a8d7b9be544e4c9e9990de2e2f8822d.png)
Thus, by reverse Power of a Point, we have that quadrilaterals 
and 
are cyclic.
Now, we move to angle chasing.
(from cyclic 
)
(from cyclic 
(from the isosceles triangle)
(from the isosceles triangle)
(from cyclic 
)
(from cyclic It is from here in which we focus on quadrilateral 
Let 
and 
Now realize that all we want to show, as given by the problem statement, is 
Thus, 
We also have:
\begin{align*}
\angle LAD&=180^\circ-(\angle ADL+\alpha)\\
&=180^\circ-((180^\circ-\gamma)+\alpha))\\
&=\gamma-\alpha.
\end{align*}
Similarly, 
Calling back to what we want to show - if we want to show that 
it suffices to show that 
and 
But this is equivalent to showing 
However, applying our Ratio Lemma to 
we have:
![\[\frac{\sin\alpha}{\sin 90^\circ}=\frac{AD}{DE}\cdot\frac{KE}{KA}.\]](http://latex.artofproblemsolving.com/b/8/4/b843b4ea3e46ec94e586a9955634861216022895.png)
Applying our Ratio Lemma to 
we have:
![\[\frac{\sin\alpha}{\sin 90^\circ}=\frac{AD}{DE}\cdot\frac{LE}{LA}.\]](http://latex.artofproblemsolving.com/f/f/1/ff13089e7954431b369d1766176212e5fd5e4ac1.png)
From the transitive property, we have:
![\[\frac{KE}{KA}=\frac{LE}{LA}.\]](http://latex.artofproblemsolving.com/3/b/6/3b62496021af6df63be223cdd9af2452071eeb74.png)
Now, realize that we already have a shared side 
and equivalent angles 
Thus, we establish a stronger bound between these two triangles - a congruence.
Thus, these two triangles are similar, and our proof is complete.
mathboy282
(cred. to v_Enhance diagram)
See Also
| 2023 USAJMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
