Difference between revisions of "User:Ddk001"
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− | 5.(Much harder) <math>\Delta ABC</math> is an isosceles triangle where <math>CB=CA</math>. Let the circumcircle of <math>\Delta ABC</math> be <math>\Omega</math>. Then, there is a point <math>E</math> and a point <math>D</math> on circle <math>\Omega</math> such that <math>AD</math> and <math>AB</math> trisects <math>\angle CAE</math> and <math>BE<AE</math>, and point <math>D</math> lies on minor arc <math> | + | 5.(Much harder) <math>\Delta ABC</math> is an isosceles triangle where <math>CB=CA</math>. Let the circumcircle of <math>\Delta ABC</math> be <math>\Omega</math>. Then, there is a point <math>E</math> and a point <math>D</math> on circle <math>\Omega</math> such that <math>AD</math> and <math>AB</math> trisects <math>\angle CAE</math> and <math>BE<AE</math>, and point <math>D</math> lies on minor arc <math>BC</math>. Point <math>F</math> is chosen on segment <math>AD</math> such that <math>CF</math> is one of the altitudes of <math>\Delta ACD</math>. Ray <math>CF</math> intersects <math>\Omega</math> at point <math>G</math> (not <math>C</math>) and is extended past <math>G</math> to point <math>I</math>, and <math>IG=AC</math>. Point <math>H</math> is also on <math>\Omega</math> and <math>AH=GI<HB</math>. Let the perpendicular bisector of <math>BC</math> and <math>AC</math> intersect at <math>O</math>. Let <math>J</math> be a point such that <math>OJ</math> is both equal to <math>OA</math> (in length) and is perpendicular to <math>IJ</math> and <math>J</math> is on the same side of <math>CI</math> as <math>A</math>. Let <math>O’</math> be the reflection of point <math>O</math> over line <math>IJ</math>. There exist a circle <math>\Omega_1</math> centered at <math>I</math> and tangent to <math>\Omega</math> at point <math>K</math>. <math>IO’</math> intersect <math>\Omega_1</math> at <math>L</math>. Now suppose <math>O’G</math> intersects <math>\Omega</math> at one distinct point, and <math>O’, G</math>, and <math>K</math> are collinear. If <math>IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2</math>, then <math>\frac{EH}{BH}</math> can be expressed in the form <math>\frac{\sqrt{b}}{a} (\sqrt{c} + d)</math>, where <math>b</math> and <math>c</math> are not divisible by the squares of any prime. Find <math>a^2+b^2+c^2+d^2+abcd</math>. |
Revision as of 18:26, 1 January 2024
See if you can solve these:
1. There is one and only one perfect square in the form
Find that perfect square.
2. Suppose there is complex values and that satisfy
Find .
3. Suppose
Find the remainder when is divided by 1000.
4. Suppose is a -degrees polynomial. The Fundamental Theorem of Algebra tells us that there are roots, say . Suppose all integers ranging from to satisfies . Also, suppose that
for an integer . If is the minimum possible value of
.
Find the number of factors of the prime in .
5.(Much harder) is an isosceles triangle where . Let the circumcircle of be . Then, there is a point and a point on circle such that and trisects and , and point lies on minor arc . Point is chosen on segment such that is one of the altitudes of . Ray intersects at point (not ) and is extended past to point , and . Point is also on and . Let the perpendicular bisector of and intersect at . Let be a point such that is both equal to (in length) and is perpendicular to and is on the same side of as . Let be the reflection of point over line . There exist a circle centered at and tangent to at point . intersect at . Now suppose intersects at one distinct point, and , and are collinear. If , then can be expressed in the form , where and are not divisible by the squares of any prime. Find .