Difference between revisions of "2023 USAJMO Problems/Problem 6"
Mathboy282 (talk | contribs) (→Solution) |
Mathboy282 (talk | contribs) (→Solution) |
||
(3 intermediate revisions by the same user not shown) | |||
Line 17: | Line 17: | ||
<cmath>\frac{\sin\theta_2’}{PC}=\frac{\sin\theta_1’}{b}=\frac{\sin C}{AP}.</cmath> | <cmath>\frac{\sin\theta_2’}{PC}=\frac{\sin\theta_1’}{b}=\frac{\sin C}{AP}.</cmath> | ||
Dividing each of the corresponding equations, we get: | Dividing each of the corresponding equations, we get: | ||
− | <cmath>\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}.</cmath> | + | <cmath>\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}.</cmath> |
Noting that <math>\sin \theta_1= \sin(180^\circ - \theta_1’)=\sin \theta_1’,</math> we have: | Noting that <math>\sin \theta_1= \sin(180^\circ - \theta_1’)=\sin \theta_1’,</math> we have: | ||
<cmath>\frac{\sin\theta_2}{\sin\theta_1}=\frac{PB}{PC}\cdot\frac{b}{c}.</cmath> | <cmath>\frac{\sin\theta_2}{\sin\theta_1}=\frac{PB}{PC}\cdot\frac{b}{c}.</cmath> | ||
Line 70: | Line 70: | ||
'''mathboy282''' | '''mathboy282''' | ||
− | [[File:2023USAJMOP6Diag.jpg | + | [[File:2023USAJMOP6Diag.jpg]] |
+ | |||
(cred. to v_Enhance diagram) | (cred. to v_Enhance diagram) |
Latest revision as of 13:41, 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:
We prove this as follows:
By the Law of Sines of we get:
By the Law of Sines of
we get:
Dividing each of the corresponding equations, we get:
Noting that
we have:
We transition into our problem.
By Power of a Point on circle we have:
By power of a point on circle
we have:
Using the transitive property, we have:
Thus, by reverse Power of a Point, we have that quadrilaterals
and
are cyclic.
Now, we move to angle chasing.
(from cyclic
)
(from the isosceles triangle)
(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:
Applying our Ratio Lemma to
we have:
From the transitive property, we have:
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.