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AMC 10 AMC 12 Guide

What is AMC 10 or 12? How to Prepare for them?

American Mathematics Contest 10/12 (AMC 10/12) is the 2nd stage of Math Olympiad Contest in the US after AMC 8. The contest is in multiple choice format and aims to develop problem solving abilities. The difficulty of the problems dynamically varies and are based on important mathematical principles. These contests have a lasting educational value.

Eligibility for AMC 10/12:

Students in grade 10 or below and under 17.5 years of age on the day of the contest can take the AMC 10. Students in grade 12 or below and under 19.5 years of age on the day of the contest can take the AMC 12.

AMC 10/12 Exam Syllabus:

The syllabus for AMC 10 include:

  • Elementary Algebra
  • Basic Geometry Knowledge like Pythagorean Theorem, area and volume formulas
  • Elementary Number Theory
  • Elementary Probability

The syllabus for AMC 10 excludes:

  • Trigonometry
  • Advanced Algebra
  • Advanced Geometry

The syllabus of AMC 12 includes high school mathematics curriculum and including all the topics mentioned above. Only calculus is excluded in the advanced topics.

AMC 10 and AMC 12 Test Format:

As mentioned earlier, the format of the tests are multiple choice type. It is a 75 minute test with 25 questions. The exams are held on two dates under the names AMC 10A and 10B for AMC 10 and AMC 12A and 12B for AMC 12.

Upcoming AMC 10 and AMC dates:

For AMC 10/12A:

Early bird Registration: through Dec. 2, 2020

Regular Registration: Dec. 3 – 19, 2020

Late Registration: Dec. 20, 2020 – Jan. 9, 2021

Exam Days: Thursday, February 4, 2021
For AMC 10/12B:

Early bird Registration: through Dec. 8, 2020

Regular Registration: Dec. 9 – Jan 9, 2020

Late Registration: Jan. 10, 2020 – 13, 2021

Exam Days: Wednesday, February 10, 2021

Books for Preparation:

Cheenta believes “It is important to completely solve at least one book per topic before going into the harder ones. You do not need to solve hard books for all topics.”

So, we prefer, one should keep “Challenges and Thrills of Pre College Mathematics” as a Common Minimum Requirement for your preparation.

How to Prepare for AMC 10 or AMC 12?

  • Utilize the resources: Here are the resources required to help you perform better in AMC 10 or AMC 12 Preparation:
  • Solve Past Papers and similar questions from similar contests like SMO (Singapore), UKMT (Junior, Senior), University of Waterloo BMC, Math Kangaroo which may serve as practice problems for AMC 10/12.
  • Take tests and Track your Tests: Tests are essential for knowing your weaknesses and strengths. Attempting quizzes and mock tests helps the students to evaluate themselves.
  • Get a coach: A coach can see and analyze your performance and guide you in the best way possible. He/She can help you strengthen your weaknesses by letting you work in the right direction.
Categories
AMC 10 AMC 12 Guide

Upcoming USA Math Competition

In this post, we are going to discuss about the upcoming USA Math Contests that would be useful for you to consider if you are looking to participate in Mathematics Competitions.

1. American Mathematics Competitions (AMC)

American Mathematics Competition (AMC) are the mathematics contests for United States students that leads them to International Mathematics Olympiad (IMO).

Eligibility:

These competitions are divided into 3 levels:

  1. AMC 8, for classes 8 or below
  2. AMC 10, for classes 10 or below
  3. AMC 12, for classes 12 or below

Higher scores in AMC 10 or 12 lead the stuedents to American Invitational Mathematics Examination (AIME) or USAMO/USAJMO.

Important Dates:

For AMC 8:

Early bird Registration: through October 1, 2020

Regular Registration: 2nd – 29th October 2020

Late Registration: 30th October – 3rd November 2020

Exam Days: Tuesday, November 10, 2020 through Monday, November 16, 2020

For AMC 10A/12A:

Early bird Registration: through Dec. 2, 2020

Regular Registration: Dec. 3 – 19, 2020

Late Registration: Dec. 20, 2020 – Jan. 9, 2021

Exam Days: Thursday, February 4, 2021
For AMC 10B/12B:

Early bird Registration: through Dec. 8, 2020

Regular Registration: Dec. 9 – Jan 9, 2020

Late Registration: Jan. 10, 2020 – 13, 2021

Exam Days: Wednesday, February 10, 2021

For AIME:

AIME I: Wednesday, March 10, 2021

AIME II: Thursday, March 18, 2021

For more info: https://www.maa.org/math-competitions

2. Perennial Math

Perennial Math has 3 types of competitions: Online Math contest, Onsite Tournament and Virtual Tournaments.

Eligibility:

Online and Onsite Tournaments are for students of Grade 3-8 and Virtual Tournaments are for students of grade 3 -12.

Important dates:

The Online Math Contest usually occurs in 2 seasons. 4 tests are held in each season. Season 1 is November through February and Season 2 is January through April.

The On Site Tournament is online for this year due to COVID 19. Here are the Ucoming Events:

  1. Birmingham AL Tournament – 24th October 2020
  2. Ohio Tournament – 14th October 2020
  3. Douglasville, GA Tournament – 15th December 2020
  4. Memphis TN Tournament – 23rd January 2020

Virtual Tournaments are Video Conferencing Tournaments in a WebEx Chat Room with a Gameshow Host. The Upcoming Virtual Tournaments are listed below:

  1. Pumpkin pi Fest (for individuals) – 17th October 2020
  2. Mathoween (for individuals) – 31st October 2020
  3. Mathgiving (for individuals) – 21st November 2020
  4. Winterchill ( for individuals) – 12th December 2020
  5. Winterdash (Grade 3-8) Team Event (for teams) – 15th – 17th December 2020
  6. Top 10 Championships (for individuals) – 19th December 2020

For more info: https://perennialmath.com/

3. SCUDEM V 2020

This fall due to COVID 19, SCUDEM V 2020 will be held in a virtual approach.

Eligibility:

SCUDEM V 2020 is an opportunity for students, in high school or undergraduate school, to engage in a differential equations modeling experience. One can register with a team of 3 or individually. A team will be formed for individuals. They can apply to participate as a student, coach or judge.

Important Dates:

  1. For Students: Registration closes 23rd October 2020
  2. For Coaches: Registration closes 15th October 2020
  3. For Judges: Registration closes 13th October 2020

For more info: https://www.simiode.org/scudem/2020

4. Rocket City Math League

The Rocket City Math League is a free, international math contest granted from Mu Alpha Theta.

Eligibility:

It  is open to all middle, high school, and two-year college students enrolled in pre-algebra through precalculus and above math courses.

Important Dates:

  1. Interschool Test: Distributed Nov 10, 2020
  2. Round One: January 12, 2021
  3. Round Two: February 16, 2021
  4. Round Three: March 9, 2021

For more info: https://mualphatheta.org/rcml-contest

5. Purple Comet! Math Meet

The Purple Comet! Math Meet is a free, annual, international, online, team, mathematics competition, where teams from 1 – 6 students compete by submitting solutions to a list of mathematics problems.

Eligibility:

It is designed for middle and high school students.

Important Dates:

Next Contest: 6 – 15th April 2021

For more info: https://purplecomet.org/

6. National Internet Math Olympiad

National Internet Math Olympiad is an online Math Contest partnered with Online Math Open (OMO). Teams of upto 4 participate in the contest who are given more than a week to answer 30 short answer questions, ranging from the early AMC to late IMO.

Eligibility:

High school students are eligible for this.

Important Dates:

Fall Contests occurs in October to November and Spring contests in March and April.

For more info: https://internetolympiad.org/contests

7. M3 Challenge

MathWorks Math Modeling (M3) Challenge helps students to to work as a team to tackle a real-world problem under time and resource constraints, akin to those faced by professional mathematicians working in industry.

Eligibility:

Students of age 16 – 19 in US participate in this competition.

Important Dates:

Registration opens in November for M3 Challenge 2021.

For more info: https://m3challenge.siam.org/

8. MathCON

MathCON is an innovative mathematics platform, including a competition series and educational resources and opportunities for students and educators, developed to enhance mathematics education.

Eligibility:

MathCON is open to all students in grades 4-12, including public, charter, private and home schooled students.

Important Dates:

Registration going on for 2021.

For more info: https://www.mathcon.org/

Categories
AMC 10 Combinatorics Math Olympiad USA Math Olympiad

Dice Problem | AMC 10A, 2014| Problem No 17

Try this beautiful Problem on Probability based on Dice from AMC 10 A, 2014. You may use sequential hints to solve the problem.

Dice Problem – AMC-10A, 2014 – Problem 17


Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

,

  • $\frac{1}{6}$
  • $\frac{13}{72}$
  • $\frac{7}{36}$
  • $\frac{5}{24}$
  • $\frac{2}{9}$

Key Concepts


combinatorics

Dice-problem

Probability

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2014 Problem-17

Check the answer here, but try the problem first

$\frac{5}{24}$

Try with Hints


First Hint

Total number of dice is \(3\) and each dice \(6\) possibility. therefore there are total $6^{3}=216$ total possible rolls. we have to find out the probability that the values shown on two of the dice sum to the value shown on the remaining die.

Without cosidering any order of the die , the possible pairs are $(1,1,2),(1,2,3),(1,3,4)$,$(1,4,5),(1,5,6),(2,2,4),(2,3,5)$,$(2,4,6),(3,3,6)$

Now can you finish the problem?

Second Hint

Clearly $(1,1,1).(2,2,4),(3,3,6)$ this will happen in $\frac{3 !}{2}=3$ way

$(1,2,3),(1,3,4)$,$(1,4,5),(1,5,6),(2,3,5)$,$(2,4,6),$this will happen in $3 !=6$ ways

Now Can you finish the Problem?

Third Hint

Therefore, total number of ways $3\times3+6\times6=45$ so that sum of the two dice will be the third dice

Therefore the required answer is $\frac{45}{216}$=$\frac{5}{24}$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Right-angled Triangle | AMC 10A, 2018 | Problem No 16

Try this beautiful Problem on Geometry based on Right-angled triangle from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Right-angled triangle – AMC-10A, 2018- Problem 16


Right triangle $A B C$ has leg lengths $A B=20$ and $B C=21$. Including $\overline{A B}$ and $\overline{B C}$, how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{A C} ?$

,

  • $5$
  • $8$
  • $12$
  • $13$
  • $15$

Key Concepts


Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2018 Problem-16

Check the answer here, but try the problem first

\(13\)

Try with Hints


First Hint

Given that \(\triangle ABC\) is a Right-angle triangle and $AB=20$ and $BC=21$. we have to find out how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$?

Let $P$ be the foot of the altitude from $B$ to $AC$. therefore \(BP\) is the shortest legth . $B P=\frac{20 \cdot 21}{29}$ which is between $14$ and $15$.

Now can you finish the problem?

Second Hint

let us assume a line segment \(BY\) with \(Y\) on \(AC\)which is starts from $A$ to $P$ . So if we move this line segment the length will be decreases and the values will be look like as \(20,…..,15\). similarly if we moving this line segment $Y$ from $P$ to $C$ hits all the integer values from $15, 16, \dots, 21$.

Now Can you finish the Problem?

Third Hint

Therefore numbers of total line segments will be \(13\)

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Categories
AMC 10 Geometry Math Olympiad USA Math Olympiad

Length of the crease | AMC 10A, 2018 | Problem No 13

Try this beautiful Problem on Geometry based on Length of the crease from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Length of the crease– AMC-10A, 2018- Problem 13


A paper triangle with sides of lengths $3,4,$ and 5 inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?

,

  • $1+\frac{1}{2} \sqrt{2}$
  • $\sqrt 3$
  • $\frac{7}{4}$
  • $\frac{15}{8}$
  • $2$

Key Concepts


Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2018 Problem-13

Check the answer here, but try the problem first

$\frac{15}{8}$

Try with Hints


First Hint

Given that ABC is a right-angle triangle shape paper. Now by the problem the point \(A\) move on point \(B\) . Therefore a crease will be create i.e \(DE\) . noe we have to find out the length of \(DE\)?

If you notice very carefully then \(DE\) is the perpendicular bisector of the line \(AB\). Therefore the \(\triangle ADE\) is Right-angle triangle. Now the side lengths of \(AC\),\(AB\),\(BC\) are given. so if we can so that the \(\triangle ADE\) \(\sim\) \(\triangle ABC\) then we can find out the side length of \(DE\)?

Now can you finish the problem?

Second Hint

In \(\triangle ABC\) and \(\triangle ADE\) we have …

\(\angle A=\angle A\)( common angle)

\(\angle C=\angle ADE\) (Right angle)

Therefore the remain angle will be equal ….

Therefore we can say that \(\triangle ADE\) \(\sim\) \(\triangle ABC\)

Now Can you finish the Problem?

Third Hint

As \(\triangle ADE\) \(\sim\) \(\triangle ABC\) therefore we can write

$\frac{B C}{A C}=\frac{D E}{A D} \Rightarrow \frac{3}{4}=\frac{D E}{\frac{5}{2}} \Rightarrow D E=\frac{15}{8}$

Therefore the length in inches of the crease is $\frac{15}{8}$

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Categories
AMC 10 Geometry Math Olympiad USA Math Olympiad

Right-angled shaped field | AMC 10A, 2018 | Problem No 23

Try this beautiful Problem on Geometry based on Right-angled shaped field from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Right-angled shaped field – AMC-10A, 2018- Problem 23


Farmer Pythagoras has a field in the shape of a right triangle. The right triangle’s legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is 2 units. What fraction of the field is planted?

,

  • $\frac{25}{27}$
  • $\frac{26}{27}$
  • $\frac{73}{75}$
  • $\frac{145}{147}$
  • $\frac{74}{75}$

Key Concepts


Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2018 Problem-23

Check the answer here, but try the problem first

\(\frac{145}{147}\)

Try with Hints


First Hint

Given that ABC is a right-angle Triangle field . Here The corner at \(B\) is shaded region which is unplanted. now we have to find out fraction of the field is planted?

Now if we join the triangle with the dotted lines then it will be divided into three triangles as shown below…

Therefore there are three triangles . Now if we can find out the area of three triangles and area of the smaller square then it will be eassy to say….

Now can you finish the problem?

Second Hint

Let \(x\) be the side length of the sqare then area will be\(x^2\)

Now area of two thin triangle will be $\frac{x(3-x)}{2}$ and $\frac{x(4-x)}{2}$

area of the other triangle will be \(\frac{1}{2}\times 5 \times 2=5\)

area of the \(\triangle ABC =\frac{1}{2}\times 3 \times 4=6\)

Now Can you finish the Problem?

Third Hint

Therefore we can say that $x^{2}+\frac{x(3-x)}{2}+\frac{x(4-x)}{2}+5=6$

\(\Rightarrow x=\frac{2}{7}\)

Therefore area of the small square will be \(\frac{4}{49}\)

Thererfore our required fraction =Area of the \(\triangle ABC\)-area of the smaller square=\(6- \frac{4}{49}\)=\(\frac{145}{147}\)

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Categories
AMC 10 Geometry Math Olympiad USA Math Olympiad

Area of region | AMC 10B, 2016| Problem No 21

Try this beautiful Geometry Problem based on area of region from AMC 10 B, 2016. You may use sequential hints to solve the problem.

Area of region– AMC-10B, 2016- Problem 21


What is the area of the region enclosed by the graph of the equation $x^{2}+y^{2}=|x|+|y| ?$

,

  • $\pi+\sqrt{2}$
  • $\pi+2$
  • $\pi+2 \sqrt{2}$
  • $2 \pi+\sqrt{2}$
  • $2 \pi+2 \sqrt{2}$

Key Concepts


Geometry

Semi circle

graph

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10B, 2016 Problem-21

Check the answer here, but try the problem first

$\pi+2$

Try with Hints


First Hint

The given equation is $x^{2}+y^{2}=|x|+|y|$. Expanding this equation we get four equation as mod exist here…

$x^2+y^2-x-y=0$…………………..(1)

$x^2+y^2+x+y=0$………………..(2)

$x^2+y^2-x+y=0$…………………(3)

$x^2+y^2+x-y=0$…………………(4)

using this four equation can you draw the figure ?

Now can you finish the problem?

Second Hint

now four equations can be written as $x^{2}-x+y^{2}-y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}+y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}-x+y^{2}+y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}-y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$ which represents four circles and they overlapping…..

The center of the four circles are $\left(\frac{1}{2}, \frac{1}{2}\right)$, $\left(\frac{-1}{2}, \frac{-1}{2}\right)$,$\left(\frac{1}{2}, \frac{-1}{2}\right)$,$\left(\frac{-1}{2}, \frac{1}{2}\right)$Now we have to find out the region union of the four circles.

Now can you finish the problem?

Third Hint

There are several ways to find the area, but note that if you connect (0,1),(1,0),(-1,0),(0,-1) to its other three respective points in the other three quadrants, you get a square of area 2 , along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2 \cdot\left(\frac{\sqrt{2}}{2}\right)^{2} \pi=\pi+2$

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Categories
AMC 10 Combinatorics Math Olympiad USA Math Olympiad

Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability based on Coin toss from AMC 10 A, 2017. You may use sequential hints to solve the problem.

Coin Toss – AMC-10A, 2017- Problem 18


Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $q-p ?$

,

  • $1$
  • $2$
  • $3$
  • $4$
  • $5$

Key Concepts


combinatorics

Coin toss

Probability

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2017 Problem-18

Check the answer here, but try the problem first

$4$

Try with Hints


First Hint

Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins.

Now can you finish the problem?

Second Hint

Let $P$ be the probability Amelia wins. Note that $P=$ chance she wins on her first turn $+$ chance she gets to her second turn $\cdot \frac{1}{3}+$ chance she gets to her third turn $\cdot \frac{1}{3} \ldots$ This can be represented by an infinite geometric series,

Therefore the value of \(P\) will be $P=\frac{\frac{1}{3}}{1-\frac{2}{3} \cdot \frac{3}{5}}=\frac{\frac{1}{3}}{1-\frac{2}{5}}=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{3} \cdot \frac{5}{3}=\frac{5}{9}$ which is of the form \(\frac{p}{q}\)

Now Can you finish the Problem?

Third Hint

Therefore \(q-p=9-5=4\)

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Categories
AMC 10 Geometry Math Olympiad USA Math Olympiad

GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry based on GCF & Rectangle from AMC 10 A, 2010. You may use sequential hints to solve the problem.

GCF & Rectangle – AMC-10A, 2016- Problem 19


In rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $B E=E F=F C$. Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q$, respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$ where the greatest common factor of $r, s,$ and $t$ is $1 .$ What is $r+s+t ?$

,

  • $7$
  • $9$
  • $12$
  • $15$
  • $20$

Key Concepts


Geometry

Rectangle

Diagonal

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2016 Problem-19

Check the answer here, but try the problem first

$20$

Try with Hints


First Hint

Given that , rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q,$ respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$. we have to find out $r+s+t ?$, where greatest common factor of \(r,s,t\) is \(1\)

Now $\triangle A P D \sim \triangle E P B$. From this relation we can find out a relation between \(DP\) and \(PB\)

Now can you finish the problem?

Second Hint

Now $\triangle A P D \sim \triangle E P B$\(\Rightarrow\) $\frac{D P}{P B}=\frac{A D}{B E}=3$ Therefore $P B=\frac{B D}{4}$.

SimIarly from the \(\triangle AQD \sim \triangle BQF\) \(\Rightarrow \)$\frac{D Q}{Q B}=\frac{3}{2}$

Therefore we can say that $D Q=\frac{3 \cdot B D}{5}$

Now can you finish the problem?

Third Hint

Therefore $r: s: t=\frac{1}{4}: \frac{2}{5}-\frac{1}{4}: \frac{3}{5}=5: 3: 12,$ so $r+s+t$=\(20\)

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Categories
AMC 10 Geometry Math Olympiad USA Math Olympiad

Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Geometry Problem based on a fly trapped inside cubical box from AMC 10 A, 2010. You may use sequential hints to solve the problem.

Fly trapped inside cubical box – AMC-10A, 2010- Problem 20


A fly trapped inside a cubical box with side length 1 meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?

,

  • $4+4 \sqrt{2}$
  • $2+4 \sqrt{2}+2 \sqrt{3}$
  • $2+3 \sqrt{2}+3 \sqrt{3}$
  • $4 \sqrt{2}+4 \sqrt{3}$
  • $3 \sqrt{2}+5 \sqrt{3}$

Key Concepts


Geometry

Cube

Diagonal

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2010 Problem-20

Check the answer here, but try the problem first

$4 \sqrt{2}+4 \sqrt{3}$

Try with Hints


First Hint

Suppose the fly starts from the point \(A\).we have to find out  the maximum possible length. The maximum possible length will be from one corner to another corner such as …..

$A \rightarrow G \rightarrow B \rightarrow H \rightarrow C \rightarrow E \rightarrow D \rightarrow F \rightarrow A$

Now can you find out this maximum path?

Now can you finish the problem?

Second Hint

Given that the side length of the cube is $1$. Therefore the diagonal \(AC\)=\(\sqrt 2\) and the diagonal \(AG=\sqrt 3\). Now we have to find out the path \(AG+GB+BH+HC+CE+ED+DF+FA\).Can you find it ?

Now can you finish the problem?

Third Hint

\(AG+GB+BH+HC+CE+ED+DF+FA\)=\(\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2\)=(\(4\sqrt3+4\sqrt2\))

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