Difference between revisions of "2014 USAJMO Problems/Problem 2"
Stevenmeow (talk | contribs) (→Solution) |
m (→Solution: ; closing lemma) |
||
(9 intermediate revisions by 6 users not shown) | |||
Line 50: | Line 50: | ||
//change O and H label positions if interfering with other lines to be added | //change O and H label positions if interfering with other lines to be added | ||
− | //further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing | + | //further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing assistance and are not to be used |
</asy> | </asy> | ||
− | + | Lemma: <math>H</math> is the reflection of <math>O</math> over the angle bisector of <math>\angle BAC</math> (henceforth 'the' reflection) | |
Proof: Let <math>H'</math> be the reflection of <math>O</math>, and let <math>B'</math> be the reflection of <math>B</math>. | Proof: Let <math>H'</math> be the reflection of <math>O</math>, and let <math>B'</math> be the reflection of <math>B</math>. | ||
Line 61: | Line 61: | ||
<math>\Delta ABB'</math> is equilateral, and <math>O</math> lies on the perpendicular bisector of <math>\overline{AB}</math> | <math>\Delta ABB'</math> is equilateral, and <math>O</math> lies on the perpendicular bisector of <math>\overline{AB}</math> | ||
− | It's well known that <math>O</math> lies strictly inside <math>\Delta ABC</math> (since it's acute), meaning that <math>\angle ABH' = \angle AB'O = 30^{\circ},</math> from which it follows that <math>\overline{BH'} \perp \ | + | It's well known that <math>O</math> lies strictly inside <math>\Delta ABC</math> (since it's acute), meaning that <math>\angle ABH' = \angle AB'O = 30^{\circ},</math> from which it follows that <math>\overline{BH'} \perp \overline{AC}</math> . Similarly, <math>\overline{CH'} \perp \overline{AB}</math>. Since <math>H'</math> lies on two altitudes, <math>H'</math> is the orthocenter, as desired. |
+ | |||
+ | End Lemma | ||
So <math>\overline{OH}</math> is perpendicular to the angle bisector of <math>\angle OAH</math>, which is the same line as the angle bisector of <math>\angle BAC</math>, meaning that <math>\Delta APQ</math> is equilateral. | So <math>\overline{OH}</math> is perpendicular to the angle bisector of <math>\angle OAH</math>, which is the same line as the angle bisector of <math>\angle BAC</math>, meaning that <math>\Delta APQ</math> is equilateral. | ||
− | Let its side length be <math>s</math>, and let <math>PH=t</math>, where <math>0 < t < s, t \neq s/2</math> because O lies strictly within <math>\angle BAC</math>, as must<math> H</math>, the reflection of <math>O</math>. Also, it's easy to show that if <math>O=H</math> in a general triangle, it's | + | Let its side length be <math>s</math>, and let <math>PH=t</math>, where <math>0 < t < s, t \neq s/2</math> because <math>O</math> lies strictly within <math>\angle BAC</math>, as must <math>H</math>, the reflection of <math>O</math>. Also, it's easy to show that if <math>O=H</math> in a general triangle, it's equilateral, and we know <math>\Delta ABC</math> is not equilateral. Hence H is not on the bisector of <math>\angle BAC \implies t \neq s/2</math>. Let <math>\overrightarrow{BH}</math> intersect <math>\overline{AC}</math> at <math>P_B</math>. |
Since <math>\Delta HP_BQ</math> and <math>BP_BA</math> are 30-60-90 triangles, <math>AB=2AP_B=2(s-QP_B)=2(s-HQ/2)=2s-HQ=2s-(s-t)=s+t</math> | Since <math>\Delta HP_BQ</math> and <math>BP_BA</math> are 30-60-90 triangles, <math>AB=2AP_B=2(s-QP_B)=2(s-HQ/2)=2s-HQ=2s-(s-t)=s+t</math> | ||
Line 74: | Line 76: | ||
The denominator equals <math>(1.5s)^2-(.5s-t)^2-s^2</math> where <math>.5s-t</math> can equal any value in <math>(-.5s, .5s)</math> except <math>0</math>. Therefore, the denominator can equal any value in <math>(s^2, 5s^2/4)</math>, and the ratio is any value in <math>\boxed{\left(\frac{4}{5},1\right)}.</math> | The denominator equals <math>(1.5s)^2-(.5s-t)^2-s^2</math> where <math>.5s-t</math> can equal any value in <math>(-.5s, .5s)</math> except <math>0</math>. Therefore, the denominator can equal any value in <math>(s^2, 5s^2/4)</math>, and the ratio is any value in <math>\boxed{\left(\frac{4}{5},1\right)}.</math> | ||
− | Note: It's easy to show that for any point <math>H</math> on <math>\ | + | Note: It's easy to show that for any point <math>H</math> on <math>\overline{PQ}</math> except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle. |
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>J</math> be the farthest point on the circumcircle of <math>\triangle{ABC}</math> from line <math>BC</math>. | ||
+ | Lemma: Line <math>AJ</math>||Line <math>OH</math> | ||
+ | Proof: Set <math>b=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}</math> and <math>c=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}</math>, and <math>a</math> on the unit circle. It is well known that <math>o=0</math> and <math>h=a+b+c</math>, so we have <math>h-o=a+b+c=a-1=a-j</math>, so <math>\dfrac{a-j}{h-o}</math> is real and thus the 2 lines are parallel. | ||
+ | |||
+ | WLOG let <math>A</math> be in the first quadrant. Clearly by the above lemma <math>OH</math> must intersect line <math>BC</math> closer to <math>B</math> than to <math>C</math>. Intersect <math>AJ</math> and <math>BC</math> at <math>D</math> and <math>OH</math> and <math>BC</math> at <math>E</math>. We clearly have <math>0 < \dfrac{\overarc{JC}-\overarc{AB}}{2} = \dfrac{120-\overarc{AC}}{2} = \angle{ADB} = \angle{OEC} < 30 = \angle{OBC}</math>, <math>OH</math> must intersect <math>AB</math>. We also have, letting the intersection of line <math>OH</math> and line <math>AC</math> be <math>Q</math>, and letting intersection of <math>OH</math> and <math>AB</math> be <math>P</math>, <math>\angle{OQA} = \angle{JAB} = \dfrac{\overarc{JB}}{2} = 60</math>. Since <math>\angle{OCA}=\angle{BCA}-\angle{BCO} < 90-30 =\angle{OQA}</math>, and <math>\angle{OQC} = 120>\angle{OAC}</math>, <math>OH</math> also intersects <math>AC</math>. We have <math>\angle{OPA}=\angle{PAJ}=60</math>, so <math>\triangle{APQ}</math> is equilateral. Letting <math>AJ=2x</math>, and letting the foot of the perpendicular from <math>O</math> to <math>AJ</math> be <math>L</math>, we have <math>OL=\sqrt{1-x^{2}}</math>, and since <math>OL</math> is an altitude of <math>\triangle{APQ}</math>, we have <math>[APQ]=\dfrac{OL^{2}\sqrt{3}}{3} = \dfrac{\sqrt{3}(1-x^{2})}{3}</math>. Letting the foot of the perpendicular from <math>A</math> to <math>OJ</math> be <math>K</math>, we have <math>\triangle{JKA}\sim \triangle{JLO}</math> by AA with ratio <math>\dfrac{JA}{JO} = 2x</math>. Therefore, <math>JK = 2x(JL) = 2x^{2}</math>. Letting <math>D</math> be the foot of the altitude from <math>J</math> to <math>BC</math>, we have <math>KD=JD-JK=\dfrac{3}{2}-2x^{2}</math>, since <math>re(B)=re(C) \implies JD=re(j)-(-\dfrac{1}{2})=\dfrac{3}{2}</math>. Thus, since <math>BC=\sqrt{3}</math> we have <math>[ABC]=\dfrac{(\dfrac{3}{2}-2x^{2})(\sqrt{3})}{2} = \dfrac{(3-4x^{2})(\sqrt{3})}{2}</math>, so <math>[PQCB]=[ABC]-[APQ]=\dfrac{(5-8x^{2})(\sqrt{3})}{12}</math>, so <math>\dfrac{[APQ]}{[PQCB]} = \dfrac{4-4x^{2}}{5-8x^{2}} = \dfrac{1}{2}+\dfrac{3}{10-16x^{2}}</math>. We have <math>x=\sin(\dfrac{JOA}{2})</math>, with <math>0<120-2\angle{ACB}=\angle{JOA}<60</math>, so <math>x</math> can be anything in the interval <math>(0, \dfrac{1}{2})</math>. Therefore, the desired range is <math>(\dfrac{4}{5}, 1)</math>. | ||
+ | |||
+ | Solution by Shaddoll | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{USAJMO box|year=2014|num-b=1|num-a=3}} |
Latest revision as of 19:40, 28 February 2023
Contents
Problem
Let be a non-equilateral, acute triangle with
, and let
and
denote the circumcenter and orthocenter of
, respectively.
(a) Prove that line intersects both segments
and
.
(b) Line intersects segments
and
at
and
, respectively. Denote by
and
the respective areas of triangle
and quadrilateral
. Determine the range of possible values for
.
Solution
Lemma: is the reflection of
over the angle bisector of
(henceforth 'the' reflection)
Proof: Let be the reflection of
, and let
be the reflection of
.
Then reflection takes to
.
is equilateral, and
lies on the perpendicular bisector of
It's well known that lies strictly inside
(since it's acute), meaning that
from which it follows that
. Similarly,
. Since
lies on two altitudes,
is the orthocenter, as desired.
End Lemma
So is perpendicular to the angle bisector of
, which is the same line as the angle bisector of
, meaning that
is equilateral.
Let its side length be , and let
, where
because
lies strictly within
, as must
, the reflection of
. Also, it's easy to show that if
in a general triangle, it's equilateral, and we know
is not equilateral. Hence H is not on the bisector of
. Let
intersect
at
.
Since and
are 30-60-90 triangles,
Similarly,
The ratio is
The denominator equals
where
can equal any value in
except
. Therefore, the denominator can equal any value in
, and the ratio is any value in
Note: It's easy to show that for any point on
except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.
Solution 2
Let be the farthest point on the circumcircle of
from line
.
Lemma: Line
||Line
Proof: Set
and
, and
on the unit circle. It is well known that
and
, so we have
, so
is real and thus the 2 lines are parallel.
WLOG let be in the first quadrant. Clearly by the above lemma
must intersect line
closer to
than to
. Intersect
and
at
and
and
at
. We clearly have
,
must intersect
. We also have, letting the intersection of line
and line
be
, and letting intersection of
and
be
,
. Since
, and
,
also intersects
. We have
, so
is equilateral. Letting
, and letting the foot of the perpendicular from
to
be
, we have
, and since
is an altitude of
, we have
. Letting the foot of the perpendicular from
to
be
, we have
by AA with ratio
. Therefore,
. Letting
be the foot of the altitude from
to
, we have
, since
. Thus, since
we have
, so
, so
. We have
, with
, so
can be anything in the interval
. Therefore, the desired range is
.
Solution by Shaddoll
See Also
2014 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |