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| − | ==User Counts==
| + | I am a aops user who likes making and doing problems, doing math, and redirecting pages (see [[Principle of Insufficient Reasons]]). I like geometry and don't like counting and probability. My number theory skill are also not bad |
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| − | If this is you first time visiting this page, change the number below by one. (Add 1, do NOT subtract 1)
| + | <br> |
| | + | __NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"> |
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| − | <math>\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{0}}}}}}</math> | + | ==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>== |
| | + | <div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div> |
| | + | <center><font size="100px">21</font></center> |
| | + | </div> |
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| − | Doesn't that look like a number on a pyramid?
| + | Credits given to [[User:Firebolt360|Firebolt360]] for inventing the box above. |
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| | ==Cool asyptote graphs== | | ==Cool asyptote graphs== |
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| | <asy>draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));</asy> | | <asy>draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));</asy> |
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| − | ==Problems I made==
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| − | See if you can solve these:
| + | ==Problems Sharing Contest== |
| | + | Here, you can post all the math problems that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start: |
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| − | 1. (Much easier) There is one and only one perfect square in the form | + | 1. There is one and only one perfect square in the form |
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| − | <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)</math> | + | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath> |
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| | where <math>p</math> and <math>q</math> are prime. Find that perfect square. | | where <math>p</math> and <math>q</math> are prime. Find that perfect square. |
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| − | 2. Suppose there is complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy
| + | <math>\textbf{Solution by cxsmi}</math> |
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| − | <math>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</math> | + | 1. We can expand the product in the expression. <math>(p^2+1)(q^2+1)-((pq)^2-pq+1) = p^2q^2+p^2+q^2+1-((pq)^2-pq+1) = p^2 + q^2 + pq</math>. Suppose this equals <math>m^2</math> for some positive integer <math>m</math>. We rewrite using the square of a binomial pattern to find that <math>m^2 = (p + q)^2 - pq</math>. Through trial and error on small values of <math>p</math> and <math>q</math>, we find that <math>p</math> and <math>q</math> must equal <math>3</math> and <math>5</math> in some order. The perfect square formed using these numbers is <math>\boxed{49}</math>. |
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| − | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
| + | Note: I will be the first to admit that this solution is somewhat lucky. |
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| − | 3. Suppose
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| − | <math>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</math> | + | 2. A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle <math>\triangle ABC</math> intersects <math>\triangle ABC</math> itself. <math>\triangle ABC</math> has leg length <math>2024</math>. The perimeter of this diamond is expressible as <math>a\sqrt{b}-c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers, and <math>c</math> is not divisible by the square of any prime. What is the remainder when <math>a + b + c</math> is divided by <math>1000</math>? |
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| − | Find the remainder when <math>\min{x}</math> is divided by 1000.
| + | <asy> |
| | + | unitsize(1inch); |
| | + | draw((0,0)--(0,2)); |
| | + | draw((0,2)--(2,0)); |
| | + | draw((2,0)--(0,0)); |
| | + | draw(circle((0.586,0.586),0.586)); |
| | + | draw((0,0)--(0,1.172),red); |
| | + | draw((0,1.172)--(1.172,1.172)); |
| | + | draw((1.172,1.172)--(1.172,0)); |
| | + | draw((1.172,0)--(0,0),red); |
| | + | draw((0,1.172)--(0.828,1.172),red); |
| | + | draw((0.828,1.172)--(1.172,0.828),red); |
| | + | draw((1.172,0.828)--(1.172,0),red); |
| | + | draw((0,0.1)--(0.1,0.1)); |
| | + | draw((0.1,0.1)--(0.1,0)); |
| | + | label("$A$",(0,2.1)); |
| | + | label("$B$",(0,-0.1)); |
| | + | label("$C$",(2,-0.1)); |
| | + | label("$2024$",(-0.2,1)); |
| | + | label("$2024$",(1,-0.2)); |
| | + | </asy> |
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| − | 4. Suppose <math>f(x)</math> is a <math>10000000010</math>-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \dots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that
| + | ==Contributions== |
| | + | [[2005 AMC 8 Problems/Problem 21]] Solution 2 |
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| − | <math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
| + | [[2022 AMC 12B Problems/Problem 25]] Solution 5 (Now it's solution 6) |
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| − | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
| + | [[2023 AMC 12B Problems/Problem 20]] Solution 3 |
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| − | <math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
| + | [[2016 AIME I Problems/Problem 10]] Solution 3 |
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| − | Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
| + | [[2017 AIME I Problems/Problem 14]] Solution 2 |
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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>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>.
| + | [[2019 AIME I Problems/Problem 15]] Solution 6 |
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| − | Someone mind making a diagram for this?
| + | [[2022 AIME II Problems/Problem 3]] Solution 3 |
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| − | ==Answer key & solution to the problems==
| + | Restored diagram for [[1994 AIME Problems/Problem 7]] |
| − | I will leave a big gap below this sentence so you won't see the answers accidentally.
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| | + | [[Divergence Theorem]] |
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| | + | [[Stokes' Theorem]] |
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| | + | [[Principle of Insufficient Reasons]] |
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| | + | ==Vandalism area== |
| | + | Here, you can add anything, delete anything, and do anything! (Don't delete this line since it's instruction and don't be inappropriate) Do not delete the see also. However, do NOT vandalize before this word (Feel free to delete this and the period that follows). |
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| | + | (ok :) :) this page is so cool!) |
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| | + | honestly i think your user page is very cool. :) |
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| | + | Hi Ddk001 [[User:zhenghua]] (Taking Oly Geo) |
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| | + | ZHENGHUA, I HAVENT SEEN YOU SINCE FOREVER!!! I'm not focusing on compitition right now so you probably won't see me in any of your classes. |
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| | + | ==See also== |
| | + | * My [[User talk:Ddk001|talk page]] |
| | + | * [[Problems Collection|My problems collection]] |
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| | + | The problems on this page are NOT copyrighted by the [http://www.maa.org Mathematical Association of America]'s [http://amc.maa.org American Mathematics Competitions]. [[File:AMC_logo.png|middle]] |
| | + | <div style="clear:both;"> |
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| − | | + | Can someone help me clear out [[Problems Collection|this page]]? |
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| − | dsf
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| − | fsd
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| − | ===Answer key===
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| − | 1. 049
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| − | 2. 170
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| − | 3. 736
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| − | 4. 011
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| − | 5. 054
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| − | ===Solutions===
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| − | ====Problem 1====
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| − | There is one and only one perfect square in the form
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| − | <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>
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| − | where <math>p</math> and <math>q</math> is prime. Find that perfect square.
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| − | ====Solution 1====
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| − | <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2 \cdot q^2 +p^2+q^2+1-p^2 \cdot q^2 +pq-1=p^2+q^2+pq</math>.
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| − | Suppose <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>.
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| − | Then, <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=(p+q)^2-pq \implies pq=(p+q)^2-n^2=(p+q-n)(p+q+n)</math>, so since <math>n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}</math>, <math>n>p,n>q</math> so <math>p+q-n</math> is less than both <math>p</math> and <math>q</math> and thus we have <math>p+q-n=1</math> and <math>p+q+n=pq</math>. Adding them gives <math>2p+2q=pq+1</math> so by [[Simon's Favorite Factoring Trick]], <math>(p-2)(q-2)=3 \implies (p,q)=(3,5)</math> in some order. Hence, <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}</math>.<math>\square</math>
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| − | | |
| − | ====Problem 2====
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| − | Suppose there are complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy
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| − | <math>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</math>
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| − | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
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| − | ====Solution 1====
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| − | To make things easier, instead of saying <math>x_i</math>, we say <math>x</math>.
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| − | Now, we have
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| − | <math>(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=\frac{1}{3}</math>.
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| − | Expanding gives
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| − | <math>x^3-(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}) \cdot x^2+(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})x-(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})=0</math>.
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| − | To make things even simpler, let
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| − | <math>a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}</math>, so that <math>x^3-ax^2+bx-c=0</math>.
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| − | Then, if <math>P_n=x_{1}^n+x_{2}^n+x_{3}^n</math>, [[Newton's Sums]] gives
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| − | <math>P_1+(-a)=0</math> <math>(1)</math>
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| − | <math>P_2+(-a) \cdot P_1+2 \cdot b=0</math> <math>(2)</math>
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| − | <math>P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0</math> <math>(3)</math>
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| − | Therefore,
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| − | <math>P_3=0-((-a) \cdot P_1+b \cdot P_1+3 \cdot (-c))</math>
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| − | <math>=a \cdot P_2-b \cdot P_1+3 \cdot c</math>
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| − | <math>=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c</math>
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| − | <math>=a(a^2-2b)-ab+3c</math>
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| − | <math>=a^3-3ab+3c</math>
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| − | Now, we plug in <math>a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}:</math>
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| − | <math>P_3=(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})^3-3(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})+3(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})</math>.
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| − | As we have done many times before, we substitute <math>x=\sqrt[3]{13},y=\sqrt[3]{53},z=\sqrt[3]{103}</math> to get
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| − | <math>P_3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{3})</math>
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| − | <math>=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3(x^2y+y^2x+x^2z+z^2x+z^2y+y^2z+3xyz)+3xyz+1</math>
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| − | <math>=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3x^2y-3y^2x-3x^2z-3z^2x-3z^2y-3y^2z-9xyz+3xyz+1</math>
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| − | <math>=x^3+y^3+z^3+1</math>
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| − | <math>=13+53+103+1</math>
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| − | <math>=\boxed{170}</math>. <math>\square</math>
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| − | Note: If you don't know [[Newton's Sums]], you can also use [[Vieta's Formulas]] to bash.
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| − | | |
| − | ====Problem 3====
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| − | Suppose
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| − | <math>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</math>
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| − | Find the remainder when <math>\min{x}</math> is divided by 1000.
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| − | ====Solution 1 (Euler's Totient Theorem)====
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| − | We first simplify <math>\cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6:</math>
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| − | <math>2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6=42^4+6 \cdot 30^6=(\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)}</math>
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| − | so
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| − | <math>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)} \equiv 1 \pmod{5}</math>
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| − | <math>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \pmod{6}</math>
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| − | <math>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 6 \cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)} \equiv 6 \pmod{7}</math>.
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| − | where the last step of all 3 congruences hold by the [[Euler's Totient Theorem]].
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| − | Hence,
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| − | <math>x \equiv 1 \pmod{5}</math>
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| − | <math>x \equiv 0 \pmod{6}</math>
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| − | <math>x \equiv 6 \pmod{7}</math>
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| − | Now, you can bash through solving linear congruences, but there is a smarter way. Notice that <math>5|x-6,6|x-6</math>, and <math>7|x-6</math>. Hence, <math>210|x-6</math>, so <math>x \equiv 6 \pmod{210}</math>. With this in mind, we proceed with finding <math>x \pmod{7!}</math>.
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| − | Notice that <math>7!=5040= \text{lcm}(144,210)</math> and that <math>x \equiv 0 \pmod{144}</math>. Therefore, we obtain the system of congruences :
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| − | <math>x \equiv 6 \pmod{210}</math>
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| − | <math>x \equiv 0 \pmod{144}</math>.
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| − | Solving yields <math>x \equiv 2\boxed{736} \pmod{7!}</math>, and we're done. <math>\square</math>
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| − | ====Problem 4====
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| − | Suppose <math>f(x)</math> is a <math>10000000010</math>-degrees polynomial. The [[Fundamental Theorem of Algebra]] tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \dots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that
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| − | <math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
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| − | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
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| − | <math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
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| − | Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
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| − | ====Solution 1====
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| − | Since all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>, we have that all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)-n=0</math>, so by the [[Factor Theorem]],
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| − | <math>n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n</math>
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| − | <math>\implies (n+1)n \dots (n-10000000008)|f(n)-n</math>.
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| − | <math>\implies f(n)=a(n+1)n \dots (n-10000000008)+n</math>
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| − | since <math>f(n)</math> is a <math>10000000010</math>-degrees polynomial, and we let <math>a</math> to be the leading coefficient of <math>f(n)</math>.
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| − | | |
| − | Also note that since <math>r_1, r_2, \dots, r_{10000000010}</math> is the roots of <math>f(n)</math>, <math>f(n)=a(n-r_1)(n-r_2) \dots (n-r_{10000000010})</math>
| |
| − | | |
| − | Now, notice that
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| − | | |
| − | <math>m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})</math>
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| − | | |
| − | <math>=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})</math>
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| − | | |
| − | <math>=\frac{f(-2)}{a}</math>
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| − | | |
| − | <math>=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}</math>
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| − | | |
| − | <math>=\frac{10000000010! \cdot a-2}{a}</math>
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| − | | |
| − | <math>=10000000010!-\frac{2}{a}</math>
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| − | | |
| − | Similarly, we have
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| − | | |
| − | <math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}</math>
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| − | | |
| − | To minimize this, we minimize <math>m</math>. The minimum <math>m</math> can get is when <math>m=10000000011</math>, in which case
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| − | | |
| − | <math>-\frac{2}{a}=10000000011!-10000000010!</math>
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| − | | |
| − | <math>=10000000011 \cdot 10000000010!-10000000010!</math>
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| − | | |
| − | <math>=10000000010 \cdot 10000000010!</math>
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| − | | |
| − | <math>\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>
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| − | | |
| − | <math>=-\frac{1}{a}</math>
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| − | | |
| − | <math>=\frac{10000000010 \cdot 10000000010}{2}</math>
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| − | | |
| − | <math>=5000000005 \cdot 10000000010!</math>
| |
| − | | |
| − | , so there is <math>\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}</math> factors of <math>999999937</math>. <math>\square</math>
| |
| − | | |
| − | ====Problem 5====
| |
| − | <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>.
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| − | | |
| − | Someone mind making a diagram for this?
| |
| − | ====Solution 1====
| |
| − | ========ss========
| |
I am a aops user who likes making and doing problems, doing math, and redirecting pages (see Principle of Insufficient Reasons). I like geometry and don't like counting and probability. My number theory skill are also not bad
User Count
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21
Credits given to Firebolt360 for inventing the box above.
Cool asyptote graphs
Asymptote is fun!
![[asy]draw((0,0)----(0,6));draw((0,-3)----(-3,3));draw((3,0)----(-3,6));draw((6,-6)----(-6,3));draw((6,0)----(-6,0));[/asy]](//latex.artofproblemsolving.com/d/6/1/d61604256e9de426afc94111a204811ff9d3cf0e.png)
![[asy]draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));[/asy]](//latex.artofproblemsolving.com/d/7/8/d78519cbbf9c22e2c4ea74ee089557db7a04e2ba.png)
Problems Sharing Contest
Here, you can post all the math problems that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start:
1. There is one and only one perfect square in the form
![\[(p^2+1)(q^2+1)-((pq)^2-pq+1)\]](//latex.artofproblemsolving.com/b/1/1/b11bf3dce41593092737df0eea5f902de3989389.png)
where 
and 
are prime. Find that perfect square.
1. We can expand the product in the expression. 
. Suppose this equals 
for some positive integer 
. We rewrite using the square of a binomial pattern to find that 
. Through trial and error on small values of and , we find that and must equal 
and 
in some order. The perfect square formed using these numbers is 
.
Note: I will be the first to admit that this solution is somewhat lucky.
2. A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle 
intersects itself. has leg length 
. The perimeter of this diamond is expressible as 
, where 
, 
, and 
are integers, and is not divisible by the square of any prime. What is the remainder when 
is divided by 
?
![[asy] unitsize(1inch); draw((0,0)--(0,2)); draw((0,2)--(2,0)); draw((2,0)--(0,0)); draw(circle((0.586,0.586),0.586)); draw((0,0)--(0,1.172),red); draw((0,1.172)--(1.172,1.172)); draw((1.172,1.172)--(1.172,0)); draw((1.172,0)--(0,0),red); draw((0,1.172)--(0.828,1.172),red); draw((0.828,1.172)--(1.172,0.828),red); draw((1.172,0.828)--(1.172,0),red); draw((0,0.1)--(0.1,0.1)); draw((0.1,0.1)--(0.1,0)); label("$A$",(0,2.1)); label("$B$",(0,-0.1)); label("$C$",(2,-0.1)); label("$2024$",(-0.2,1)); label("$2024$",(1,-0.2)); [/asy]](//latex.artofproblemsolving.com/c/f/4/cf4b2b8da31fb32285d5b573ffbb35e870afea4c.png)
Contributions
2005 AMC 8 Problems/Problem 21 Solution 2
2022 AMC 12B Problems/Problem 25 Solution 5 (Now it's solution 6)
2023 AMC 12B Problems/Problem 20 Solution 3
2016 AIME I Problems/Problem 10 Solution 3
2017 AIME I Problems/Problem 14 Solution 2
2019 AIME I Problems/Problem 15 Solution 6
2022 AIME II Problems/Problem 3 Solution 3
Restored diagram for 1994 AIME Problems/Problem 7
Divergence Theorem
Stokes' Theorem
Principle of Insufficient Reasons
Vandalism area
Here, you can add anything, delete anything, and do anything! (Don't delete this line since it's instruction and don't be inappropriate) Do not delete the see also. However, do NOT vandalize before this word (Feel free to delete this and the period that follows).
(ok :) :) this page is so cool!)
honestly i think your user page is very cool. :)
Hi Ddk001 User:zhenghua (Taking Oly Geo)
ZHENGHUA, I HAVENT SEEN YOU SINCE FOREVER!!! I'm not focusing on compitition right now so you probably won't see me in any of your classes.
See also
The problems on this page are NOT copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
![[asy]draw((0,0)----(0,6));draw((0,-3)----(-3,3));draw((3,0)----(-3,6));draw((6,-6)----(-6,3));draw((6,0)----(-6,0));[/asy]](http://latex.artofproblemsolving.com/d/6/1/d61604256e9de426afc94111a204811ff9d3cf0e.png)
![[asy]draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));[/asy]](http://latex.artofproblemsolving.com/d/7/8/d78519cbbf9c22e2c4ea74ee089557db7a04e2ba.png)
![\[(p^2+1)(q^2+1)-((pq)^2-pq+1)\]](http://latex.artofproblemsolving.com/b/1/1/b11bf3dce41593092737df0eea5f902de3989389.png)
and 
are prime. Find that perfect square.
. Suppose this equals 
for some positive integer 
. We rewrite using the square of a binomial pattern to find that 
. Through trial and error on small values of 
and 
in some order. The perfect square formed using these numbers is 
.
intersects 
. The perimeter of this diamond is expressible as 
, where 
, 
, and 
are integers, and 
is divided by 
?
![[asy] unitsize(1inch); draw((0,0)--(0,2)); draw((0,2)--(2,0)); draw((2,0)--(0,0)); draw(circle((0.586,0.586),0.586)); draw((0,0)--(0,1.172),red); draw((0,1.172)--(1.172,1.172)); draw((1.172,1.172)--(1.172,0)); draw((1.172,0)--(0,0),red); draw((0,1.172)--(0.828,1.172),red); draw((0.828,1.172)--(1.172,0.828),red); draw((1.172,0.828)--(1.172,0),red); draw((0,0.1)--(0.1,0.1)); draw((0.1,0.1)--(0.1,0)); label("$A$",(0,2.1)); label("$B$",(0,-0.1)); label("$C$",(2,-0.1)); label("$2024$",(-0.2,1)); label("$2024$",(1,-0.2)); [/asy]](http://latex.artofproblemsolving.com/c/f/4/cf4b2b8da31fb32285d5b573ffbb35e870afea4c.png)
