Difference between revisions of "2023 USAMO Problems/Problem 1"
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In an acute triangle <math>ABC</math>, let <math>M</math> be the midpoint of <math>\overline{BC}</math>. Let <math>P</math> be the foot of the perpendicular from <math>C</math> to <math>AM</math>. Suppose that the circumcircle of triangle <math>ABP</math> intersects line <math>BC</math> at two distinct points <math>B</math> and <math>Q</math>. Let <math>N</math> be the midpoint of <math>\overline{AQ}</math>. Prove that <math>NB=NC</math>. | In an acute triangle <math>ABC</math>, let <math>M</math> be the midpoint of <math>\overline{BC}</math>. Let <math>P</math> be the foot of the perpendicular from <math>C</math> to <math>AM</math>. Suppose that the circumcircle of triangle <math>ABP</math> intersects line <math>BC</math> at two distinct points <math>B</math> and <math>Q</math>. Let <math>N</math> be the midpoint of <math>\overline{AQ}</math>. Prove that <math>NB=NC</math>. | ||
== Solution 1 == | == Solution 1 == | ||
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Let <math>X</math> be the foot from <math>A</math> to <math>\overline{BC}</math>. By definition, <math>\angle AXM = \angle MPC = 90^{\circ}</math>. Thus, <math>\triangle AXM \sim \triangle MPC</math>, and <math>\triangle BMP \sim \triangle AMQ</math>. | Let <math>X</math> be the foot from <math>A</math> to <math>\overline{BC}</math>. By definition, <math>\angle AXM = \angle MPC = 90^{\circ}</math>. Thus, <math>\triangle AXM \sim \triangle MPC</math>, and <math>\triangle BMP \sim \triangle AMQ</math>. | ||
− | From this, we have <math>\frac{MP}{MX} = \frac{ | + | From this, we have <math>\frac{MP}{MX} = \frac{MC}{MA} = \frac{MP}{MQ} = \frac{MA}{MB}</math>, as <math>MC=MB</math>. Thus, <math>M</math> is also the midpoint of <math>XQ</math>. |
− | Now, <math>NB = NC</math> | + | Now, <math>NB = NC</math> if <math>N</math> lies on the perpendicular bisector of <math>\overline{BC}</math>. As <math>N</math> lies on the perpendicular bisector of <math>\overline{XQ}</math>, which is also the perpendicular bisector of <math>\overline{BC}</math> (as <math>M</math> is also the midpoint of <math>XQ</math>), we are done. |
− | ~ Martin2001, | + | ~ Martin2001 |
+ | |||
+ | ==Solution 2 == | ||
+ | |||
+ | We are going to use barycentric coordinates on <math>\triangle ABC</math>. Let <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, <math>C=(0,0,1)</math>, and <math>a=BC</math>, <math>b=CA</math>, <math>c=AB</math>. We have <math>M=\left(0,\frac{1}{2},\frac{1}{2}\right)</math> and <math>P=(x:1:1)</math> so <math>\overrightarrow{CP}=\left(\frac{x}{x+2},\frac{1}{x+2},\frac{1}{x+2}-1\right)</math> and <math>\overrightarrow{AM}=\left(-1,\frac{1}{2},\frac{1}{2}\right)</math>. Since <math>\overleftrightarrow{CP}\perp\overleftrightarrow{AM}</math>, it follows that | ||
+ | <cmath>\begin{align*} | ||
+ | a^2\left(\frac{1}{2}\cdot\frac{1}{x+2}+\frac{1}{2}\left(\frac{1}{x+2}-1\right)\right)+b^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\left(\frac{1}{x+2}-1\right)\right)\\ | ||
+ | +c^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\frac{1}{x+2}\right)=0. | ||
+ | \end{align*}</cmath> | ||
+ | Solving this gives | ||
+ | <cmath>\[ | ||
+ | x=\frac{2b^2-2c^2}{a^2-3b^2-c^2} | ||
+ | \]</cmath> | ||
+ | so | ||
+ | <cmath>\[ | ||
+ | P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right). | ||
+ | \]</cmath> | ||
+ | The equation for <math>(ABP)</math> is | ||
+ | <cmath>\[ | ||
+ | -a^2yz-b^2zx-c^2xy+ux+vy+wz=0. | ||
+ | \]</cmath> | ||
+ | Plugging in <math>A</math> and <math>B</math> gives <math>u=v=0</math>. Plugging in <math>P</math> gives | ||
+ | <cmath>\begin{align*} | ||
+ | -a^2\left(\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right)^2-b^2\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\\ | ||
+ | -c^2\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}+w\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}=0 | ||
+ | \end{align*}</cmath> | ||
+ | so | ||
+ | <cmath>\[ | ||
+ | w=\frac{2b^4-2c^4+a^4-3a^2b^2-a^2c^2}{2a^2-4b^2-4c^2}=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2}. | ||
+ | \]</cmath> | ||
+ | Now let <math>Q=(0,t,1-t)</math> where | ||
+ | <cmath>\begin{align*} | ||
+ | -a^2t(1-t)+w(1-t)&=0\\ | ||
+ | \implies t&=\frac{w}{a^2} | ||
+ | \end{align*}</cmath> | ||
+ | so <math>Q=\left(0,\frac{w}{a^2},1-\frac{w}{a^2}\right)</math>. It follows that <math>N=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)</math>. It suffices to prove that <math>\overleftrightarrow{ON}\perp\overleftrightarrow{BC}</math>. Setting <math>\overrightarrow{O}=0</math>, we get <math>\overrightarrow{N}=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)</math>. Furthermore we have <math>\overrightarrow{CB}=(0,1,-1)</math> so it suffices to prove that | ||
+ | <cmath>\begin{align*} | ||
+ | a^2\left(-\frac{w}{2a^2}+\frac{1}{2}-\frac{u}{2a^2}\right)+b^2\left(-\frac{1}{2}\right)+c^2\left(\frac{1}{2}\right)=0\\ | ||
+ | \implies w=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2} | ||
+ | \end{align*}</cmath> | ||
+ | which is valid. <math>\square</math> | ||
+ | |||
+ | ~KevinYang2.71 | ||
+ | |||
+ | ==See also== | ||
+ | {{USAMO box|year=2023|before=First Problem|num-a=2|n=I}} |
Latest revision as of 01:55, 6 February 2024
In an acute triangle , let
be the midpoint of
. Let
be the foot of the perpendicular from
to
. Suppose that the circumcircle of triangle
intersects line
at two distinct points
and
. Let
be the midpoint of
. Prove that
.
Solution 1
Let be the foot from
to
. By definition,
. Thus,
, and
.
From this, we have , as
. Thus,
is also the midpoint of
.
Now, if
lies on the perpendicular bisector of
. As
lies on the perpendicular bisector of
, which is also the perpendicular bisector of
(as
is also the midpoint of
), we are done.
~ Martin2001
Solution 2
We are going to use barycentric coordinates on . Let
,
,
, and
,
,
. We have
and
so
and
. Since
, it follows that
Solving this gives
so
The equation for
is
Plugging in
and
gives
. Plugging in
gives
so
Now let
where
so
. It follows that
. It suffices to prove that
. Setting
, we get
. Furthermore we have
so it suffices to prove that
which is valid.
~KevinYang2.71
See also
2023 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |