MathFun4Kids

MATHCOUNTS Exercise – Convolution of Non-zero Squares

Posted on March 1, 2021 by kevin

A four by four grid of unit squares contains squares of various sizes (1 by 1 through 4 by 4), each of which are formed entirely from squares in the grid. In each of the 16 unit squares, write the number of squares that contain it. For instance, the middle numbers in the top row are 6s because they are each contained in one $1\times 1$ square, two $2\times 2$, two $3\times 3$, and one $4\times 4$.
(a) What is the sum of all sixteen numbers written in this grid?
(b) What about the same problem with a $10\times 10$ grid?

Click here for the solution.

Posted in Algebra, Combinatorics | Comments Off

MATHCOUNTS Exercise – 22

Posted on January 1, 2021 by kevin

Point $E$ and $F$ are inside the square $ABCD$, with $DE=10$, $EF=6$, $BF=4$, and $\angle{DEF}=\angle{BFE}=90^\circ$. Find the area of the square $ABCD$. Click here for the solutions.

Solution 1: Draw line $DB$, intersecting $EF$ at $G$. We have $\triangle{DEG}\sim\triangle{BFG}$.

$\therefore \dfrac{FG}{EG}=\dfrac{BF}{DE}=\dfrac{4}{10}$

$\because FG+EG=EF=6$

$\therefore FG=\dfrac{12}{7}$, and $EG=\dfrac{30}{7}$

$DG=\sqrt{DE^2+EG^2}=\sqrt{10^2+(\dfrac{30}{7})^2}=\dfrac{10}{7}\sqrt{58}$

$BG=\sqrt{BF^2+FG^2}=\sqrt{4^2+(\dfrac{12}{7})^2}=\dfrac{4}{7}\sqrt{58}$

$\therefore BD=DG+BG=\dfrac{10}{7}\sqrt{58}+\dfrac{4}{7}\sqrt{58}=2\sqrt{58}$

$\therefore$ Area of $ABCD=\dfrac{BD^2}{2}=\dfrac{(2\sqrt{58})^2}{2}=\boxed{116}$

Solution 2: Draw line $DB$. Point $G$ is on $DE$ so that $DG=6$. Extend $BF$ to $H$ so that $BH=6$. Draw line $GH$ intersecting $DB$ at $I$. We have $\triangle{DGI}\cong\triangle{BHI}$.

$\therefore GI=HI=\dfrac{1}{2}GH=\dfrac{1}{2}{EF}=3$

$\therefore DI^2=BI^2=DG^2+GI^2=6^2+3^2=58$

$\therefore$ Area of $ABCD=2\cdot DI^2=2\cdot 58=\boxed{116}$

Solution 3: Let point $G$ on $DE$ so that $GE=6$. Extend $BF$ to $H$ so that $FH=6$. Draw line $AH$ and $CE$.

$\therefore GE=EF=FH=GH=6$, and $EFGH$ is a square.

$\because B$, $F$, and $H$ are co-liner, $D$, $G,$ and $E$ are co-liner

$\therefore A$, $H$, and $G$ are co-linear, $C$, $E$, and $F$ are co-liner

$\therefore \triangle{ABH}\cong\triangle{BCF}\cong\triangle{CDE}\cong\triangle{DAG}$

$\therefore$ Area of $ABCD=EFHG+4\cdot\triangle{CDE}=EF^2+4\cdot\dfrac{DE\cdot CE}{2}=6^2+4\cdot\dfrac{4\cdot 10}{2}=\boxed{116}$

Solution 4: Extend line $DE$ to $G$ so that $EG=4$. Draw line $BD$ and $BG$.

$\because EG=BF=4$ and $\angle{DEF}=\angle{BFE}=90^\circ$

$\therefore BGEF$ is a rectangle

$\therefore \triangle{BDG}$ is a right triangle

$\therefore BD^2=BG^2+DG^2=EF^2+(DE+EG)^2=6^2+(10+4)^2=232$

$\therefore$ Area of $ABCD=\dfrac{1}{2}\cdot BD^2=\dfrac{1}{2}\cdot 232=\boxed{116}$

Generalization: Point $E$ and $F$ are inside square $ABCD$, with $DE=a$, $EF=b$, $BF=c$, and $\angle{DEF}=\angle{BFE}=90^\circ$, then the area of the square $ABCD$ is $$ABCD=\boxed{\dfrac{(a+c)^2+b^2}{2}}$$

Extra: Andrew started at the entry point $D$, which is one corner of a square shaped maze $ABCD$, with the exit point $B$ directly across the diagonal of the square. The following is his track record:

He entered the maze at point $D$ and walked $20$ feet first along the path.
He changed his direction $90^\circ$ clock-wise and walked another $20$ feet.
He changed his direction $90^\circ$ counter-clock-wise and walked another $20$ feet
He repeated step 2 and step 3 two more times.
He repeated step 3 once again, then step 2 twice.
He changed his direction $90^\circ$ counter-clock-wise and walked $40$ feet to reach the exit point $B$.

Based on the above track record, calculate the area of the maze. Click here for the solution.

Solution: Based on the track record, the net distance Andrew walked parallel to the initial path direction $DE$ is $140$ feet, and the net distance perpendicular to the initial path direction $DE$ is $60$ feet, as shown the diagram in the right. The length of the diagonal $BD$ of the square $ABCD$ is:

$BD=\sqrt{PD^2+PB^2}=\sqrt{140^2+60^2}=20\sqrt{58}$

Therefore, the area of $ABCD$ is:

$ABCD=\dfrac{BD^2}{2}=\dfrac{(20\sqrt{58})^2}{2}=\boxed{11600}$

Posted in Geometry, MATHCOUNTS | Comments Off

Sum and Product of All Factors

Posted on December 18, 2020 by kevin

For a nature number $$n=p_{1}^{q_{1}}\times p_{2}^{q_{2}}\times \cdots \times p_{m}^{q_{m}}$$ where $p_{i} (1 \le i \le m) $ are unique prime numbers.

The total number of positive factors of $n$ is $$f=(q_{1}+1)\times(q_{2}+1)\times\cdots\times(q_{m}+1)$$ The sum of all positive factors of $n$ is $$\sum_{d|n}d=\dfrac{p_{1}^{q_1 +1}-1}{ p_1 -1}\times \dfrac{p_{2}^{q_{2}+1}-1}{ p_{2}-1}\times \cdots \times\dfrac{p_{m}^{q_{m}+1}-1}{p_{m} -1}$$ The product of all positive factors of $n$ is $$\prod_{d|n}d=n^{\frac{f}{2}}$$ i.e. the $\dfrac{f}{2}th$ power of $n$, where $f$ is the total number of positive factors of $n$.

Question 1: A number is perfect if the sum of all positive factors is twice of the number. What is the sum of the first $4$ perfect numbers found by Euclid?

Question 2: A natural number is multiplicative perfect if the square of the number is the product of all positive factors. What is the sum of the $3$ smallest multiplicative perfect numbers greater than $1$?

Question 3: The sum of all positive factors of a natural number is 403. What is this number?

Question 4: What is the largest natural number less than $1000$ such that the product of all positive factors is the number raised to the $12$th power?

Posted in Algebra, Combinatorics | Comments Off

Cheat Sheet for Distributing $k$ Balls into $n$ Boxes

Posted on November 19, 2020 by kevin

Distribution of		Restriction
$k\text{ Balls}$	$n\text{ Boxes}$	None	$\le 1$	$\ge 1$	$=1$
$distinct$	$distinct$	$$n^k$$	$$(n)_k$$	$$n!S(k,n)$$	$$n!\text{ or }0$$
$identical$	$distinct$	$${n+k-1}\choose k$$	$$n\choose k$$	$${k-1}\choose{n-1}$$	$$1\text{ or }0$$
$distinct$	$identical$	$$\sum_{i=1}^{n}S(k,i)$$	$$1\text{ or }0$$	$$S(k,n)$$	$$1\text{ or }0$$
$identical$	$identical$	$$\sum_{i=1}^{n}P(k,i)$$	$$1\text{ or }0$$	$$P(k,n)$$	$$1\text{ or }0$$

$(n)_k=n(n-1)(n-2)…(n-k+1)=k!{n\choose k}$

$S(k, n)$ is a Stirling number of the second kind:

$$S(k,n)=\dfrac{1}{n!}\sum_{i=0}^{n}(-1)^{i}{{n}\choose{i}}(n-i)^k$$

$P(k, n)$ is the number of partitions of $k$ into $n$ parts

Posted in Combinatorics | Comments Off

Geometry Fold Problems

Posted on October 16, 2020 by kevin

A unit equilateral $\triangle{ABC}$ is folded over line $DE$, forming a quadrilateral $BCDE$, with $A$ touching $BC$ at $A’$, and $\triangle{BA’E}$ is a right triangle. The area of $BCDE$ is __________. Answer: $\dfrac{17\sqrt{3}-27}{8}\approx 0.305608$

A quarter-circle $ABC$, with its center at $A$ and its radius as $1$, is folded over $BD$, with its center touching the arc $BC$ at $A’$. The area of the resulting figure is __________. Answer: $\dfrac{3\pi-2\sqrt{3}}{12}\approx 0.49672$

Two opposite vertices $A$ and $C$ of a unit square $ABCD$ are folded toward the diagonal $BD$, forming a kite-shaped area $BEDF$. The area of the kite is __________.

Point $A$ of a unit square $ABCD$ is folded toward the diagonal $BD$ at point $A’$. Then, point $D$ is folded over $AC$, touching point $B$, resulting in a quadrilateral $BCFE$. The area of $BCFE$ is __________.

Point $A$ is the center of a quarter circle $ABC$ with radius as $1$. Point $D$ is the midpoint of $AB$. Point $C$ is folded over $EF$, touching point $D$, The length of $EF$ is __________. Answer: $\dfrac{2\sqrt{71}-3}{40}\sqrt{5} \approx 0.774367$

Point $A$ of unit square $ABCD$ is folded over $FG$, touching point $E$, which is the midpoint of $CD$. The length of $FG$ is __________. Answer: $\dfrac{\sqrt{5}}{2}$

A semi-circle with $AB$ as its diameter, point $O$ as its center, and its radius as $1$. The circle is folded over chord $BC$ and intersecting the diameter at $O$. The area bounded by $\overparen{AC}$, $AB$ and $BC$ is __________. Answer: $\dfrac{\pi}{6}+\dfrac{\sqrt{3}}{4}$

A semi-circle with $AB$ as its diameter, point $O$ as its center, and its radius as $5$. The circle is folded over chord $BD$ and intersecting the diameter at $C$, and $\dfrac{AC}{BC}=\dfrac{1}{2}$. The length of $BD$ is __________ (PUMaC 2010).

A semi-circle with $AB$ as its diameter, point $O$ as its center, and its radius as $2$. The circle is folded over chord $CD$, tangent with the $AB$ at $E$, with $\dfrac{BE}{AE}=\dfrac{1}{3}$. The length of $CD$ is __________.

$\triangle{ABC}$ is a right triangle with $AB=BC=12$, and $\angle{ABC}=90^\circ$. $D$ is the midpoint of $BC$. Point $A$ is folded over $EF$ to touch point $D$. The length of $EF$ is __________.

$\triangle{ABC}$ is a right triangle with $AB=3$, $BC=4$, and $\angle{ABC}=90^\circ$. $D$ is the midpoint of $BC$. Point $A$ is folded over $EF$ to touch point $D$. The length of $EF$ in the simplest form is $\dfrac{a}{b}\sqrt{c}$, where $a$, $b$, and $c$ are integers. The value of $a+b+c$ is __________.

$\triangle{ABC}$ is a right triangle with $AB=BC=8$, and $\angle{ABC}=90^\circ$. $D$ is a point on $AC$ so that $\dfrac{CD}{AC}=\dfrac{1}{3}$. Point $B$ is folded over $EF$ to touch point $D$, resulting a concave polygon $A’EFCD$. The length of $EF$ is __________.

A pentagon was folded from a square of paper, as shown in the figure. At first the edges $BC$ and $DC$ were folded to the diagonal $AC$, so that the corners $B$ and $D$ lie on the diagonal and then the resulting shape was folded so that the vertex $C$ coincided with the vertex A. The value of the angle indicated by the question mark is __________.

A paper strip is folded three times as shown. If $\alpha=70^\circ$, then $\beta=$ __________ .

In a unit equalaterial $\triangle{ABC}$, point $A$ is folded to the point $D$ on $BC$ as shown, resulting in the crease $EF$ with $E$ on $AB$ and $F$ on $AC$. If $FD\perp BC$:

The value of $\angle{AED}$ is __________.
The length of $CD$ is __________.
The ratio of the areas of $\triangle{AEF}$ and $\triangle{ABC}$ is __________.

Posted in Geometry | Comments Off

MATHCOUNTS Exercise – Convolution of Non-zero Squares

MATHCOUNTS Exercise – 22

Sum and Product of All Factors

Cheat Sheet for Distributing $k$ Balls into $n$ Boxes

Geometry Fold Problems

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