Difference between revisions of "2014 USAJMO Problems/Problem 2"
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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 <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 <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>. | + | 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> |
Revision as of 17:00, 18 August 2015
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
Claim: 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.
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 |