Difference between revisions of "2014 USAJMO Problems/Problem 2"
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//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>. | ||
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<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. |
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> | ||
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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 21:03, 23 December 2016
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.
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 |