Difference between revisions of "2019 USAJMO Problems/Problem 3"
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− | By | + | By monotonicity, we can see that the point <math>P</math> is unique. Therefore, if we find another point <math>P'</math> with all the same properties as <math>P</math>, then <math>P = P'</math> |
Part 1) Let <math>N</math> be a point on <math>\overline{AB}</math> such that <math>AN\cdot AB = AD^2</math>, and <math>BN \cdot AB = BC^2</math>. Obviously <math>N</math> exists because adding the two equations gives <math>AN\cdot AB + BN \cdot AB = AD^2 + BC^2 = AB^2</math>, which is the problem statement. Notice that converse PoP gives<cmath>AN\cdot AB = AD^2 \implies \bigtriangleup ADN \sim \bigtriangleup ABD</cmath><cmath>BN\cdot AB = AD^2 \implies \bigtriangleup CBN \sim \bigtriangleup ABC</cmath>Therefore, <math>\angle AND = \angle ADB = \angle ACB = \angle CNB</math>, so <math>N</math> does indeed satisfy all the conditions <math>P</math> does, so <math>N = P</math>. Hence, <math>\bigtriangleup ADP \sim \bigtriangleup ABD</math> and <math>\bigtriangleup CBP \sim\bigtriangleup ABC</math>. | Part 1) Let <math>N</math> be a point on <math>\overline{AB}</math> such that <math>AN\cdot AB = AD^2</math>, and <math>BN \cdot AB = BC^2</math>. Obviously <math>N</math> exists because adding the two equations gives <math>AN\cdot AB + BN \cdot AB = AD^2 + BC^2 = AB^2</math>, which is the problem statement. Notice that converse PoP gives<cmath>AN\cdot AB = AD^2 \implies \bigtriangleup ADN \sim \bigtriangleup ABD</cmath><cmath>BN\cdot AB = AD^2 \implies \bigtriangleup CBN \sim \bigtriangleup ABC</cmath>Therefore, <math>\angle AND = \angle ADB = \angle ACB = \angle CNB</math>, so <math>N</math> does indeed satisfy all the conditions <math>P</math> does, so <math>N = P</math>. Hence, <math>\bigtriangleup ADP \sim \bigtriangleup ABD</math> and <math>\bigtriangleup CBP \sim\bigtriangleup ABC</math>. |
Revision as of 20:11, 12 April 2021
Contents
Problem
Let be a cyclic quadrilateral satisfying . The diagonals of intersect at . Let be a point on side satisfying . Show that line bisects .
Solution 1
Let . Also, let be the midpoint of .
Note that only one point satisfies the given angle condition. With this in mind, construct with the following properties:
Claim:
Proof:
The conditions imply the similarities and whence as desired.
Claim: is a symmedian in
Proof:
We have
as desired.
Since is the isogonal conjugate of , . However implies that is the midpoint of from similar triangles, so we are done.
~sriraamster
Solution 2
By monotonicity, we can see that the point is unique. Therefore, if we find another point with all the same properties as , then
Part 1) Let be a point on such that , and . Obviously exists because adding the two equations gives , which is the problem statement. Notice that converse PoP givesTherefore, , so does indeed satisfy all the conditions does, so . Hence, and .
Part 2) Define as the midpoint of . Furthermore, create a point such that and . Obviously must be a parallelogram. Now we set up for Jacobi's. The problem already gives us that , which is good for starters. Furthermore, tells us thatThis gives us our second needed angle equivalence. Lastly, will givewhich is our last necessary angle equivalence to apply Jacobi's. Finally, applying Jacobi's tells us that , , and are concurrent , , collinear. Additionally, since parallelogram diagonals bisect each other, , , and are collinear, so finally we obtain that , , and are collinear, as desired.
-jj_ca888
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |