Difference between revisions of "2016 USAJMO Problems/Problem 5"
(→Solution 3) |
m (→Solution 6) |
||
(12 intermediate revisions by 5 users not shown) | |||
Line 6: | Line 6: | ||
== Solution 1== | == Solution 1== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair O=(0,0); | ||
+ | pair A=dir(110); | ||
+ | pair B=dir(-29); | ||
+ | pair C=dir(209); | ||
+ | pair H=foot(A,B,C); | ||
+ | pair P=foot(H,A,B); | ||
+ | pair Q=foot(H,A,C); | ||
+ | draw(A--B--C--A--H--P); | ||
+ | draw(circle(O,1)); | ||
+ | draw(Q--H); | ||
+ | |||
+ | dot("$A$", A, dir(A)); | ||
+ | dot("$B$", B, dir(B)); | ||
+ | dot("$C$", C, dir(C)); | ||
+ | dot("$H$", H, S); | ||
+ | dot("$P$", P, NE); | ||
+ | dot("$Q$", Q, NW); | ||
+ | dot("$O$", O, S); | ||
+ | </asy> | ||
+ | |||
It is well-known that <math>AH\cdot 2AO=AB\cdot AC</math> (just use similar triangles or standard area formulas). Then by Power of a Point, | It is well-known that <math>AH\cdot 2AO=AB\cdot AC</math> (just use similar triangles or standard area formulas). Then by Power of a Point, | ||
<cmath>AP\cdot AB=AH^2=AQ\cdot AC</cmath> Consider the transformation <math>X\mapsto \Psi(X)</math> which dilates <math>X</math> from <math>A</math> by a factor of <math>\dfrac{AB}{AQ}=\dfrac{AC}{AP}</math> and reflects about the <math>A</math>-angle bisector. Then <math>\Psi(O)</math> clearly lies on <math>AH</math>, and its distance from <math>A</math> is <cmath>AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH</cmath> so <math>\Psi(O)=H</math>, hence we conclude that <math>O,P,Q</math> are collinear, as desired. | <cmath>AP\cdot AB=AH^2=AQ\cdot AC</cmath> Consider the transformation <math>X\mapsto \Psi(X)</math> which dilates <math>X</math> from <math>A</math> by a factor of <math>\dfrac{AB}{AQ}=\dfrac{AC}{AP}</math> and reflects about the <math>A</math>-angle bisector. Then <math>\Psi(O)</math> clearly lies on <math>AH</math>, and its distance from <math>A</math> is <cmath>AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH</cmath> so <math>\Psi(O)=H</math>, hence we conclude that <math>O,P,Q</math> are collinear, as desired. | ||
Line 34: | Line 56: | ||
Clearly, <math>AO = R.</math> Since <math>AH^2 = 2\cdot AO^2,</math> we have <math>AH = \sqrt{2}R.</math> Thus, <math>AH\cdot AO = \sqrt{2}R^2.</math> | Clearly, <math>AO = R.</math> Since <math>AH^2 = 2\cdot AO^2,</math> we have <math>AH = \sqrt{2}R.</math> Thus, <math>AH\cdot AO = \sqrt{2}R^2.</math> | ||
− | Note that <math>AH = b\sin\gamma = c\sin\beta.</math> | + | Note that <math>AH = b\sin\gamma = c\sin\beta.</math> The Extended Law of Sines states that: |
− | |||
− | |||
− | The Extended Law of Sines states that: | ||
<cmath>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.</cmath> | <cmath>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.</cmath> | ||
− | Therefore, <math>AP = \frac{b^2 c}{4R^2}</ | + | Therefore, <math>AH = \frac{bc}{2R} = \sqrt{2}R.</math> Thus, <math>bc = \sqrt{2}R^2.</math> |
+ | |||
+ | Since <math>\angle PHA = \beta</math> and <math>\angle QHA = \gamma,</math> we have: | ||
+ | <cmath>AP = AH\sin\beta = c\sin^2\beta = \frac{b^2 c}{4R^2}</cmath> | ||
+ | <cmath>AQ = AH\sin\gamma = b\sin^2\gamma = \frac{bc^2}{4R^2}</cmath> | ||
+ | It follows that: | ||
<cmath>AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.</cmath> | <cmath>AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.</cmath> | ||
We see that <math>AP\cdot AQ = AH\cdot AO.</math> | We see that <math>AP\cdot AQ = AH\cdot AO.</math> | ||
Line 47: | Line 71: | ||
Rearranging <math>AP\cdot AQ = AH\cdot AO,</math> we get <math>\frac{AP}{AO} = \frac{AO}{AH}.</math> We also have <math>\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,</math> so <math>\triangle PAO\sim\triangle HAQ</math> by SAS similarity. Thus, <math>\angle AOP = \angle AQH,</math> so <math>\angle AOP</math> is a right angle. | Rearranging <math>AP\cdot AQ = AH\cdot AO,</math> we get <math>\frac{AP}{AO} = \frac{AO}{AH}.</math> We also have <math>\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,</math> so <math>\triangle PAO\sim\triangle HAQ</math> by SAS similarity. Thus, <math>\angle AOP = \angle AQH,</math> so <math>\angle AOP</math> is a right angle. | ||
− | Since <math>\angle AOP</math> and <math>\angle AOQ</math> are both right angles, we get <math>\angle POQ = \pi,</math> so we conclude that <math>P, O, Q</math> are collinear, | + | Since <math>\angle AOP</math> and <math>\angle AOQ</math> are both right angles, we get <math>\angle POQ = \pi,</math> so we conclude that <math>P, O, Q</math> are collinear, and we are done. (We also obtain the extra interesting fact that <math>AO\perp PQ.</math>) |
+ | |||
+ | == Solution 4== | ||
+ | Draw the altitude from <math>O</math> to <math>AB</math>, and let the foot of this altitude be <math>D</math>. | ||
+ | |||
+ | Then, by the Right Triangle Altitude Theorem on triangle <math>AHB</math>, we have: <math>AB\cdot AP=AH^{2}</math>. | ||
+ | |||
+ | Since <math>OD</math> is the perpendicular bisector of <math>AB</math>, <math>2\cdot AD = AB</math>. | ||
+ | |||
+ | Substituting this into our previous equation gives <math>2\cdot AD \cdot AP = AH^{2}</math>, which equals <math>2\cdot AO^{2}</math> by the problem condition. | ||
+ | |||
+ | Thus, <math>2\cdot AD\cdot AP = 2\cdot AO^{2} \implies AD\cdot AP = AO^{2}</math>. | ||
+ | |||
+ | Again, by the Right Triangle Altitude Theorem, angle <math>AOP</math> is right. | ||
+ | |||
+ | By dropping an altitude from <math>O</math> to <math>AC</math> and using the same method, we can find that angle <math>AOQ</math> is right. Since <math>\angle AOP=\angle AOQ=90</math>, <math>P</math>, <math>O</math>, <math>Q</math> are collinear and we are done. | ||
+ | |||
+ | ~champion999 | ||
+ | |||
+ | == Solution 5== | ||
+ | We use complex numbers. Let lower case letters represent their respective upper case points, with <math>|a| = |b| = |c| = 1</math>. Spamming the foot from point to segment formula, we obtain <cmath>h = \dfrac{1}{2}\left(a+b+c-\dfrac{bc}{a}\right),</cmath> <cmath>p = \dfrac{1}{2}(h+a+b-ab\overline{h}) = \dfrac{1}{4}\left(2a+2b+c-\dfrac{bc}{a}-\dfrac{ab}{c}+\dfrac{a^2}{c}\right),</cmath> and <cmath>q = \dfrac{1}{2}(h+a+c-ac\overline{h}) = \dfrac{1}{4}\left(2a+b+2c-\dfrac{bc}{a}-\dfrac{ac}{b}+\dfrac{a^2}{b}\right).</cmath> We now simplify the given length condition: <cmath>\begin{align*} AH^2 &= 2AO^2 \\ \implies (h-a)\overline{(h-a)} &= 2 \\ \implies \dfrac{1}{2}\left(-a+b+c-\dfrac{bc}{a}\right) \cdot \dfrac{1}{2}\left(-\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{a}{bc}\right) &= 2 \\ \implies (ab+ac-bc-a^2)^2 &= 8a^2bc \\ \implies (a-b)^2(a-c)^2 &= 8a^2bc.\end{align*}</cmath> We would like to show that <math>P</math>, <math>O</math>, <math>Q</math> are collinear, or <cmath>\begin{vmatrix} p & \overline{p} & 1 \\ 0 & 0 & 1 \\ q & \overline{q} & 1 \end{vmatrix} = \overline{p}q - p\overline{q} = 0.</cmath> After some factoring (or expanding) that takes about 15 minutes, this eventually reduces to <cmath>\overline{p}q - p\overline{q} = 0 \iff (a-b)(a-c)(b-c)((a-b)^2(a-c)^2 - 8a^2bc) = 0,</cmath> which is true. <math>\square</math> | ||
+ | |||
+ | -MP8148 | ||
+ | |||
+ | ==Solution 6== | ||
+ | Claim: <math>\triangle AOQ \sim \triangle AHB</math> | ||
+ | |||
+ | Proof: We compute the area of <math>\triangle ABC</math> using two methods. Let <math>R=AO</math> be the circumradius of <math>\triangle ABC</math>. | ||
+ | |||
+ | First, by extended law of sines, <math>BC=2R \sin \angle BAC</math>. We are also given that <math>AH= R \sqrt{2}</math>. | ||
+ | <math>AH \perp BC</math>, so <cmath>[ABC]=\frac{2R \sin \angle BAC \cdot R \sqrt{2}}{2}=R^2 \sin \angle BAC \sqrt{2}.</cmath> | ||
+ | |||
+ | Second, we compute the area using <math>[ABC]=\frac{1}{2} \cdot AB \cdot AC \sin \angle BAC</math>. | ||
+ | |||
+ | Equating these two expressions for the area of <math>\triangle ABC</math> and reducing, we get <cmath>AB \cdot AC = 2 R^2 \sqrt{2}.</cmath> | ||
+ | But <math>2R^2 =2 AO^2 = AH^2</math>, so <math>AB \cdot AC = AH^2 \sqrt{2}</math> and <math>\bf{AB =\frac{AH^2 \sqrt{2}}{AC}}</math>. | ||
+ | |||
+ | Since <math>\angle AQH=\angle AHC =90^\circ</math>, and both triangles share the angle <math>\angle HAC</math>, <math>\triangle AHC \sim \triangle AQH</math>. This tells us <math>\frac{AQ}{AH}=\frac{AH}{AC}</math> or <math>AQ =\frac{AH^2}{AC}</math>. | ||
+ | |||
+ | Substituting into the bolded equation, we get <math>AB=AQ \sqrt{2}</math>. | ||
+ | |||
+ | The original length condition can be written as <math>AH= AO \sqrt{2}</math>. | ||
+ | |||
+ | We also have <cmath>\angle OAQ= \angle OAC= 90- \angle ABC = \angle HAB.</cmath> | ||
+ | |||
+ | Therefore, by SAS similarity, <math>\triangle AOQ \sim \triangle AHB</math>. | ||
+ | ----------------------------------------- | ||
+ | We can prove analogously that <math>\triangle AOP \sim \triangle AHC</math>. | ||
+ | We now have <math>\angle AOQ=\angle AHB =90^\circ</math> and <math>\angle AOP=\angle AHC =90^\circ</math>. | ||
+ | This implies that <math>\angle POQ=\angle POA +\angle QOA=180^\circ</math> which tells us <math>P,O,</math> and <math>Q</math> are collinear, as desired.<math>\square</math> | ||
+ | -vvluo | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:15, 14 June 2020
Contents
Problem
Let be an acute triangle, with
as its circumcenter. Point
is the foot of the perpendicular from
to line
, and points
and
are the feet of the perpendiculars from
to the lines
and
, respectively.
Given that prove that the points
and
are collinear.
Solution 1
It is well-known that (just use similar triangles or standard area formulas). Then by Power of a Point,
Consider the transformation
which dilates
from
by a factor of
and reflects about the
-angle bisector. Then
clearly lies on
, and its distance from
is
so
, hence we conclude that
are collinear, as desired.
Solution 2
We will use barycentric coordinates with respect to The given condition is equivalent to
Note that
Therefore, we must show that
Expanding, we must prove
Let such that
The left side is equal to
The right side is equal to
which is equivalent to the left hand side. Therefore, the determinant is
and
are collinear.
Solution 3
For convenience, let denote the lengths of segments
respectively, and let
denote the measures of
respectively. Let
denote the circumradius of
Since the central angle subtends the same arc as the inscribed angle
on the circumcircle of
we have
Note that
so
Thus,
Similarly, one can show that
(One could probably cite this as well-known, but I have proved it here just in case.)
Clearly, Since
we have
Thus,
Note that The Extended Law of Sines states that:
Therefore,
Thus,
Since and
we have:
It follows that:
We see that
Rearranging we get
We also have
so
by SAS similarity. Thus,
so
is a right angle.
Rearranging we get
We also have
so
by SAS similarity. Thus,
so
is a right angle.
Since and
are both right angles, we get
so we conclude that
are collinear, and we are done. (We also obtain the extra interesting fact that
)
Solution 4
Draw the altitude from to
, and let the foot of this altitude be
.
Then, by the Right Triangle Altitude Theorem on triangle , we have:
.
Since is the perpendicular bisector of
,
.
Substituting this into our previous equation gives , which equals
by the problem condition.
Thus, .
Again, by the Right Triangle Altitude Theorem, angle is right.
By dropping an altitude from to
and using the same method, we can find that angle
is right. Since
,
,
,
are collinear and we are done.
~champion999
Solution 5
We use complex numbers. Let lower case letters represent their respective upper case points, with . Spamming the foot from point to segment formula, we obtain
and
We now simplify the given length condition:
We would like to show that
,
,
are collinear, or
After some factoring (or expanding) that takes about 15 minutes, this eventually reduces to
which is true.
-MP8148
Solution 6
Claim:
Proof: We compute the area of using two methods. Let
be the circumradius of
.
First, by extended law of sines, . We are also given that
.
, so
Second, we compute the area using .
Equating these two expressions for the area of and reducing, we get
But
, so
and
.
Since , and both triangles share the angle
,
. This tells us
or
.
Substituting into the bolded equation, we get .
The original length condition can be written as .
We also have
Therefore, by SAS similarity, .
We can prove analogously that .
We now have
and
.
This implies that
which tells us
and
are collinear, as desired.
-vvluo
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2016 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |