GYanggg

Circle's general form is \(ax^2 + by^2 + cx + dy + e = 0.\)

We want to rewrite \(x^2 + y^2 = 2x + 2y\) to its standard form. 

\(x^2 + y^2 = 2x + 2y\\ x^2-2x+y^2-2y=0\\ x^2-2x+1+y^2-2y+1=2\\ (x-1)^2+(y-1)^2=2\)

The circle has a center of \((1,1)\) and a radius of \(\sqrt2. \)

Largest x value on the circle graph is farthest right, \(\boxed{1+\sqrt2}\)

If there is any part you don't understand, please message me

The first step in order to tackle this problem is to draw another square enclosing square DEFG.

First, we prove that \(WXYZ\) is actually a square: 

\( \because\overline{ED}=\overline{DG}=\overline{GF}=\overline{FE}\because\angle{Z}=\angle{Y}=\angle{YXW}=\angle{ZWX}\because\angle{DEZ}+\angle{WEF}=90º,\angle{DEZ}+\angle{ZDE}=90º\Rightarrow\angle{ZDE}=\angle{WEF}.\)

\(\text{Using the same reasoning, we get:} \angle{ZDE}=\angle{WEF}=\angle{DGY}=\angle{GFX}.\)
\(\therefore\text{By AAS congruency:} \triangle{ZDE}\cong\triangle{YGD}\cong\triangle{XFG}\cong\triangle{WEF}.\)

From this, we get

\(\overline{ZE}+\overline{EW}=\overline{ZD}+\overline{DY}=\overline{YG}+\overline{GX}=\overline{FX}+\overline{FW}, \)which simplifies to \(\overline{ZW}=\overline{ZY}=\overline{YX}=\overline{XW}.\)

Therefore \(WXYZ \) is a square.

Since \(\triangle ABC\) is equilateral,\( \angle B=60º. \because \triangle BEW\) is a 30-60-90 triangle, \(\frac{\overline{EW}}{\overline{BW}}=\sqrt3.  \text{Same goes with } \triangle GXC, \frac{\overline{GX}}{\overline{XC}}=\sqrt3.\)
\(\text{If }\overline{EW}=x \text{ and } \overline{GX}=y,\text{ we get }\overline{BW}=\frac{x}{\sqrt3} \text{ and } \overline{XC}=\frac{y}{\sqrt3}.\)

If the equilateral triangle's side length is \(a, a=\overline{BW}+\overline{WF}+\overline{FX}+\overline{XC}=\frac{x}{\sqrt3}+y+x+\frac{y}{\sqrt3}.\)

After simplifying, we get \(x+y=\frac{3-\sqrt3}{2}a.\)

\(\because \overline{WX}=x+y \therefore x+y=\overline{DH}.\)

Since in any case, \(\overline{DH}=\frac{3-\sqrt3}{2}a, \) the length remains consistent.

Expanding the terms, 

\((x^2+2xh+h^2)+(y^2+2yh+y^2)=(a^2+2ab+b^2) \)

Using Pythagorean Theorem,

\(x^2+h^2=a^2\\ y^2+h^2=b^2\)

We could subsitute the values in, and rewrite the equation

\(a^2+2xh+b^2+2yh=a^2+2ab+b^2\\ 2xh+2yh=2ab\\ h(x+y)=ab \)

\([ABC]=\frac12 AB\cdot CH = \frac12 BC \cdot AC\\ \frac12 (x+y)h=\frac12 ab\\ h(x+y)=ab\)

I hope this helped,

Gavin

Expanding the terms, 

\((x^2+2xh+h^2)+(y^2+2yh+y^2)=(a^2+2ab+b^2) \)

Using Pythagorean Theorem,

\(x^2+h^2=a^2\\ y^2+h^2=b^2\)

We could subsitute the values in, and rewrite the equation

\(a^2+2xh+b^2+2yh=a^2+2ab+b^2\\ 2xh+2yh=2ab\\ h(x+y)=ab \)

\([ABC]=\frac12 AB\cdot CH = \frac12 BC \cdot AC\\ \frac12 (x+y)h=\frac12 ab\\ h(x+y)=ab\)

I hope this helped,

Gavin

\(\because \angle DHE=\angle FHJ \) (vertical angles) 

\(\because EH=HF\) (midpoint)

\(\because \overline{DE}\parallel\overline{FJ} \Rightarrow \angle EDJ =\angle FJD\)  (alternating interior angles)

\(\therefore \triangle HDE\cong\triangle HJF\) (AAS Congruency) 

\(\therefore [HDE]=[HJF] \Rightarrow [CDHF]+[HDE]=[CDHF]+[HJF]\)

\([CDJ]=[CDEF]\)

\(\boxed{36}\)

I hope this helped,

Gavin. 

1. The expression can be re-written as \((1+\frac1{1000})^{1000}\). Since Euler's constant, \(e\), is equilavent to \((1+\frac1n)^n\) when \(n\) approches infinity. Therefore, the expression is approimately \(e=2.717\cdots\). 

2. \(\frac92\) (I'll add the solution if I have time). 

(a)

Let \(M\) be the midpoint of \(\overline{BC}\), connect \(A\) to \(M\). \(\overline {AM}\) must pass through \(G\) since \(G\) is the centroid and \(\overline {AM}\) is a median of \(\triangle ABC\). In addition, connect \(D\) to \(M\), \(\overline {DM}\) must pass through \(H\) since \(H\) is the centroid and \(\overline {DM}\) is the median of \(\triangle BCD\).

\(\because \frac{MG}{MA}=\frac13=\frac{MH}{MD} \text{ and } \angle AMD = \angle AMD \therefore\)  by SAS similarity, \(\triangle MGH \sim \triangle MAD\). Therefore\( \overline {GH} \parallel \overline {AD} \text{ and } AD=3GH.\) Using the same method to prove \(\triangle NKJ \sim \triangle NAD\), we can conclude \(\overline {KJ} \parallel \overline {AD} \text{ and } AD=3KJ. \because AD=3GH \text{ and } AD = 3KJ \therefore GH=KJ\), similarly  \(\because \overline {KJ} \parallel \overline {AD} \text{ and }\overline {GH} \parallel \overline {AD} \therefore \overline {KJ} \parallel\overline {GH}\).

The same method is used to prove every pair of opposite sides in the hexagon are parallel and equal in length.

(b)

Let the parallel line of \(\overline{HI}\) through \(G\) and the parallel line of \(\overline{GH}\) through \(I\) intersect at . \(\because \overline{GH} \parallel \overline {OI} \text { and } \overline {GO} \parallel \overline{HI} \therefore GHIO\) is a parallelogram. Then, we can conclude \([GHI]=[GOI]\). \(\because \overline{GH} \parallel \overline {OI} \text{ and } \overline {KJ} \parallel\overline {GH} \therefore \overline {OI} \parallel \overline {KJ} \because GH = OI \text { and } GH = KJ \therefore OI = KJ\).

From this, we can conclude \(OIJK\) is also a parallelogram, which means \([IOK]=[IJK]\). \(\because \overline{HI} \parallel \overline {GO} \text{ and } \overline {HI} \parallel\overline {LK} \therefore \overline {GO} \parallel \overline {LK} \because HI = GO \text { and } HI=LK \therefore GO = LK\). From this, we can conclude \(GOKL\) is also a parallelogram, which means \([GOK]=[GLK]\). Therefore, \([GHIJKL]=2[GIK]\). Using the same method, we can conclude \([GHIJKL]=2[LHJ]\), which means that \([LHJ]=[GIK]\).

I hope this helped,

Gavin. 

1. 

Let \(a = x^2 - 10x - 29\)

\(\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0.\\  (a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0 \\ -64a + 40 \times 16 = 0, \Rightarrow a = 10. \\\)

\(10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x + 3)\) The positive root is \(\boxed{13}.\)

2. 

https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_5

I hope this helped,

Gavin