As shown in the figure, $ABCD$ is a square, $E$ is the mid-point of $AB$. The circle with its center at $H$ is tangent with $AD$, $AE$ and $DE$. The circle with its center at $F$ is tangent with $BC$, $BE$ and $DE$. The circle with its center at $G$ is tangent with $BC$, $CD$ and $DE$. Prove that $\triangle{FGH}$ is a right triangle. Click here for the proof.
Proof: Extend $DE$ and $CB$ so that they intersect at $K$. Draw diagonals of $ABCD$ so that they intersect at $O$. Let $M$ and $N$ are tangent points of circle $F$ on $DE$ and $AB$ respectively, we have $FM\perp DE$ and $FN\perp AB$. Additionally, without loss of generality, we assume that $ABCD$ is a unit square. It is obvious that $F$ is on $BD$, $G$ and $H$ are on $AC$.
Let the radius of circle $F$, $G$, $H$ as $f$, $g$, $h$ respectively. We have ${DF}^2={FM}^2+{DM}^2$
We have $$((1-f)\sqrt{2})^2=(\dfrac{\sqrt{5}-1}{2}+f)^2+f^2$$
Solve the above equation, we have $$f=\dfrac{\sqrt{5}-1}{4}$$
Since the perimeter of $\triangle{DAE}$ is $\dfrac{3+\sqrt{5}}{2}$, and its area is $\dfrac{1}{4}$. Therefore, $$h=\dfrac{2\cdot\dfrac{1}{4}}{\dfrac{3+\sqrt{5}}{2}}=\dfrac{3-\sqrt{5}}{4}$$
Since $\triangle{DAE}\sim \triangle{KCD}$, and $CD=2{AE}$, therefore $$g=2h=\dfrac{3-\sqrt{5}}{2}$$
It is easy to show that ${FO}^2={GO}\cdot{HO}$, therefore, $\triangle{FGH}$ is a right triangle.
Proof 1: Without loss of generality, assume that $ABCDEF$ is unit regular hexgon with its center at $(0,0)$, $A$ at $(-\dfrac{1}{2},-\dfrac{\sqrt{3}}{2})$, $B$ at $(\dfrac{1}{2},-\dfrac{\sqrt{3}}{2})$, $C$ at $(1,0)$, etc. Then the liner equations for line $AB$, $BC$, $CD$, and $EF$ are:
$y+\dfrac{\sqrt{3}}{2}=0$
$\sqrt{3}x-y-\sqrt{3}=0$
$\sqrt{3}x+y-\sqrt{3}=0$
$\sqrt{3}x-y+\sqrt{3}=0$
Let $O$ at $(a, b)$. Denote $O(XY)$ as the distance from $O$ to line $XY$. The distances from $O$ to $AB$, $BC$, $CD$ and $EF$ are:
Because $MB+MC=BC=1$, we have $$\dfrac{2\sqrt{3}}{3}(OL-\dfrac{1}{2}OM)+\dfrac{2\sqrt{3}}{3}(ON-\dfrac{1}{2}OM)=1\tag{1}$$
Simplifying $(1)$, we have $$OL+ON-OM=\dfrac{\sqrt{3}}{2} \tag{2}$$
Because $BC\parallel EF$, $OK\perp EF$ and $OM\perp BC$, $M$, $O$ and $M$ are co-linear and $$OK+OM=KM=\sqrt{3}$$
Applying $(2)$, We have $$OK=\sqrt{3}-OM=2(OL+ON-OM)-OM=2\cdot OL+2\cdot ON-3\cdot OM$$
which implies $[\triangle{OEF}]=2[\triangle{OAB}]+2[\triangle{OCD}]-3[\triangle{OBC}]$.
Proof 3: Without loss of generality, assume that $ABCDEF$ is unit regular hexgon.
Let $GL$ pass $O$ and $GL\perp AB$, $KM$ pass $O$ and $KM\perp BC$, $HN$ pass $O$ and $HN\perp CD$. As $ABCDEF$ is a regular hexgon, $AB\parallel DE$, $BC\parallel EF$, and $CD\parallel FA$. Therefore $GL\perp DE$, $KM\perp EF$, and $HN\perp FA$.
Extend $AB$, $CD$, and $EF$ to form equilateral $\triangle{TUV}$, with its side length as $3$. Because $OK\perp VT$, $OL\perp TU$, and $ON\perp UV$, we have $$OK+OL+ON=\dfrac{3\sqrt{3}}{2}\tag{1}$$
as $OK$, $OL$ and $ON$ are heights of $\triangle{OEF}$, $\triangle{OAB}$ and $\triangle{OCD}$ respectively and their sum is equal to the height of equilateral $\triangle{TUV}$. Similarly, $$OG+OH+OM=\dfrac{3\sqrt{3}}{2}\tag{2}$$
In right triangle ABC, with AB = 44 cm and BC = 33 cm, point D lies on side BC so that BD:DC = 2:1. If vertex A is folded onto point D to create quadrilateral BCEF, as shown, what is the area of triangle CDE? Click here for the solution.
Solution 1: Obviously, $\triangle{ABC}$ is a 3-4-5 triangle with $AC=55$.
Because $BD:DC=2:1$, and $BC=33$, we have $BD=22$, $CD=11$.
Let $BF=x$, $AF=DF=AB-BF=44-x$. Because $\triangle{BDF}$ is a right triangle, according to Pythagorean Theorem, we have
Solve the above equation, we have $x=\dfrac{33}{2}$. Therefore $BF=\dfrac{33}{2}$, $AF=\dfrac{55}{2}$.
Draw line $AD$ intersecting $EF$ at $G$, and extend $EF$ and $CB$ intersecting at $H$, as shown below. We have $AG\perp EF$ as $\triangle{DEF}$ is a reflection of $\triangle{AEF}$ along line $EF$. $\triangle{AGF}$ is a right triangle.
Therefore, $\triangle{AGF}\sim\triangle{ABD}\sim\triangle{HBF}$. We have $$\dfrac{HB}{BF}=\dfrac{AB}{BD}=\dfrac{44}{22}=2$$ Therefore $HB=2BF=33$, and $CH=HB + BC=66$.
According to Menelaus’s Theorem, we have $$\dfrac{CH}{HB}\cdot\dfrac{BF}{FA}\cdot\dfrac{AE}{EC}=1$$i.e.
Because $$AE+EC=AC=55\tag{3}$$ Solving (2) and (3), we have: $AE=25$, $EC=30$.
Because $\triangle{DEF}$ is a reflection of $\triangle{AEF}$, $DE=AE=25$. So the lengths of three sides of $\triangle{CDE}$ are $a=11$, $b=25$, and $c=30$. According to Heron’s formula, we have $$s=\dfrac{a+b+c}{2}=33$$ and the area of $\triangle{CDE}$ is
Because $\triangle{DEF}$ is a reflection of $\triangle{AEF}$, $AG\perp GE$. Therefore, $\triangle{AGE}\sim\triangle{AHD}$ and $AG=\dfrac{1}{2}AD=11\sqrt{5}$.
The area of $\triangle{ACD}$ is $$[\triangle{ACD}]=\dfrac{1}{2}AB\cdot CD=\dfrac{1}{2}\times 44\times 11=242$$
Because $[\triangle{AHD}]=\dfrac{1}{2}AC\cdot DH$, we have $$242=\dfrac{1}{2}\times 55\cdot DH$$
Therefore $DH=\dfrac{44}{5}$.
Because $\triangle{AHD}$ is a right triangle, we have $$AH=\sqrt{AD^2-DH^2}=\sqrt{(22\sqrt{5})^2-(\dfrac{44}{5})^2}=22\sqrt{5-\dfrac{4}{25}}=22\times\dfrac{11}{5}=\dfrac{242}{5}$$
Because $\triangle{AGE}\sim\triangle{AHD}$, we have $$\dfrac{GE}{AG}=\dfrac{DH}{AH}=\dfrac{\dfrac{44}{5}}{\dfrac{242}{5}}=\dfrac{2}{11}$$
As $\angle{BAE}=60^\circ$, $\triangle{ABE}$ is equilateral. Therefore $\angle{AEB}=60^\circ$
Because $\angle{ADB}=30^\circ$, $\angle{AEB}=2\angle{ADB}$, and $EA=EB$, $D$ must be on the circle with its center at $E$ and its radius as $EB$. Therefore $AE=BE=DE$. And $\triangle{AED}$ is isosceles.
We have $\angle{CED}=\angle{DAE}+\angle{ADE}=10^\circ+10^\circ=20^\circ$. Therefore $\triangle{CDE}$ is isosceles. $CD=DE$. As $DE=BE=AE=AB$, we have $AB=CD$.
Let $a$, $b$, $c$ be real numbers with $|a-b|=3$, $|b-c|=4$, and $|c-d|=5$. What is the sum of all possible values of $|a-d|$?
What is the greatest integer less than or equal to $$\dfrac{5^{200}+3^{200}}{5^{197}+3^{197}}$$
How many negative integers can be written in the form of $$a_4\cdot 5^4+a_3\cdot 5^3 +a_2\cdot 5^2+a_2\cdot 5^1+a_0\cdot 5^0$$ where $a_i \in \{-2,-1,0,1,2\}$ for $0\le i \le 4$.
What is the sum of all real numbers $x$ for which the median of the numbers $4$, $6$, $8$, $17$, and $x$ is equal to the mean of those five numbers?
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The number $2023$ is expressed in the form $$2023=\dfrac{a_1!a_2!…a_m!}{b_1!b_2!…b_n!}$$ where $a_1\ge a_2\ge … \ge a_m$ and $b_1\ge b_2\ge …\ge b_n$ are positive integers and $a_1+b_1$ is as small as possible. What is $|a_1-b_1|$?
Two increasing sequences of positive integers have different first terms. Each sequence has the property that each term beginning with the third term is the sum of the previous two terms, and the 7th term of each sequence is $N$. What is the smallest possible value of $N$?
In a certain card game, a player is dealt a hand of 13 cards from a deck of 52 distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $A35013559A00$. What is the digit of $A$?
In the following equation, each letter represents uniquely a different digit in base 10: $$XY\times YX=X20Y$$ Find the value of $XY$.
A digital clock is malfunctioning as it always skips digit 3. For example, if the clock is displaying $04:29$, the next minute its display would show $04:40$. The clock was reset as $00:00$ at midnight. Later that day, its display shows the current time as $12:56$. What should be the correct time?
