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+ | <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>. | ||
+ | |||
+ | Someone mind making a diagram for this? | ||
+ | ====Solution 5==== |
Revision as of 14:42, 3 January 2024
Contents
User Counts
If this is you first time visiting this page, change the number below by one. (Add 1, do NOT subtract 1)
Doesn't that look like a number on a pyramid?
Cool asyptote graphs
Asymptote is fun!
Problems I made
See if you can solve these:
1. (Much easier) There is one and only one perfect square in the form
where and
are prime. 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 positive integral 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
.
Someone mind making a diagram for this?
Answer key & solution to the problems
I will leave a big gap below this sentence so you won't see the answers accidentally.
dsf
fsd
Answer key
1. 049
2. 170
3. 736
4. 011
5. 054
Solutions
Problem 1
There is one and only one perfect square in the form
where and
is prime. Find that perfect square.
Solution 1
.
Suppose
.
Then,
, so since
,
so
is less than both
and
and thus we have
and
. Adding them gives
so by Simon's Favorite Factoring Trick,
in some order. Hence,
.
Problem 2
Suppose there are complex values and
that satisfy
Find .
Solution 1
To make things easier, instead of saying , we say
.
Now, we have
.
Expanding gives
.
To make things even simpler, let
, so that
.
Then, if , Newton's Sums gives
Therefore,
Now, we plug in
.
As we have done many times before, we substitute to get
.
Note: If you don't know Newton's Sums, you can also use Vieta's Formulas to bash.
Problem 3
Suppose
Find the remainder when is divided by 1000.
Solution 1 (Euler's Totient Theorem)
We first simplify
so
.
where the last step of all 3 congruences hold by the Euler's Totient Theorem. Hence,
Now, you can bash through solving linear congruences, but there is a smarter way. Notice that , and
. Hence,
, so
. With this in mind, we proceed with finding
.
Notice that and that
. Therefore, we obtain the system of congruences :
.
Solving yields , and we're done.
Problem 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 positive integral value of
.
Find the number of factors of the prime in
.
Solution 1
Since all integers ranging from
to
satisfies
, we have that all integers
ranging from
to
satisfies
, so by the Factor Theorem,
.
since is a
-degrees polynomial, and we let
to be the leading coefficient of
.
Also note that since is the roots of
,
Now, notice that
Similarly, we have
To minimize this, we minimize . The minimum
can get is when
, in which case
, so there is factors of
.
Problem 5
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
.
Someone mind making a diagram for this?