Difference between revisions of "2019 USAJMO Problems/Problem 3"
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==Problem== | ==Problem== | ||
− | + | Let <math>ABCD</math> be a cyclic quadrilateral satisfying <math>AD^2+BC^2=AB^2</math>. The diagonals of <math>ABCD</math> intersect at <math>E</math>. Let <math>P</math> be a point on side <math>\overline{AB}</math> satisfying <math>\angle APD=\angle BPC</math>. Show that line <math>PE</math> bisects <math>\overline{CD}</math>. | |
− | ==Solution== | + | ==Solution 1== |
+ | |||
+ | Let <math>PE \cap DC = M</math>. Also, let <math>N</math> be the midpoint of <math>AB</math>. | ||
+ | |||
+ | Note that only one point <math>P</math> satisfies the given angle condition. With this in mind, construct <math>P'</math> with the following properties: | ||
+ | |||
+ | <math>AP' \cdot AB = AD^2 \quad \text{and} \quad BP' \cdot AB = CD^2</math> | ||
+ | |||
+ | Claim: <math>P = P'</math> | ||
+ | |||
+ | Proof: | ||
+ | |||
+ | The conditions imply the similarities <math>ADP \sim ABD</math> and <math>BCP \sim BAC</math> whence <math>\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB</math> as desired. <math>\square</math> | ||
+ | |||
+ | Claim: <math>PE</math> is a symmedian in <math>AEB</math> | ||
+ | |||
+ | Proof: | ||
+ | |||
+ | We have | ||
+ | |||
+ | <cmath>AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB</cmath> | ||
+ | <cmath>\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}</cmath> | ||
+ | <cmath>\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1</cmath> | ||
+ | <cmath>\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}</cmath> | ||
+ | <cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath> | ||
+ | |||
+ | as desired. <math>\blacksquare</math> | ||
+ | |||
+ | Since <math>P</math> is the isogonal conjugate of <math>N</math>, <math>\measuredangle PEA = \measuredangle MEC = \measuredangle BEN</math>. However <math>\measuredangle MEC = \measuredangle BEN</math> implies that <math>M</math> is the midpoint of <math>CD</math> from similar triangles, so we are done. <math>\blacksquare</math> | ||
+ | |||
+ | ~sriraamster | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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 2) Define <math>G</math> as the midpoint of <math>CD</math>. Furthermore, create a point <math>X</math> such that <math>DX || EC</math> and <math>CX || ED</math>. Obviously <math>XCED</math> must be a parallelogram. Now we set up for Jacobi's. The problem already gives us that <math>\angle APD = \angle CPB</math>, which is good for starters. Furthermore, <math>\bigtriangleup ADP \sim \bigtriangleup ABD</math> tells us that<cmath>\angle ADP = \angle ABD = \angle ACD = \angle XDC</cmath>This gives us our second needed angle equivalence. Lastly, <math>\bigtriangleup CBP \sim\bigtriangleup ABC</math> will give<cmath>\angle BCP = \angle BAC = \angle BDC = \angle XCD</cmath>which is our last necessary angle equivalence to apply Jacobi's. Finally, applying Jacobi's tells us that <math>AC</math>, <math>BD</math>, and <math>XP</math> are concurrent <math>\implies</math> <math>X</math>, <math>E</math>, <math>P</math> collinear. Additionally, since parallelogram diagonals bisect each other, <math>X</math>, <math>G</math>, and <math>E</math> are collinear, so finally we obtain that <math>P</math>, <math>E</math>, and <math>G</math> are collinear, as desired. | ||
+ | |||
+ | -jj_ca888 | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:13, 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 gives
Therefore,
, 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 that
This gives us our second needed angle equivalence. Lastly,
will give
which 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 |