Difference between revisions of "2024 AMC 12A Problems/Problem 5"

Revision as of 21:41, 20 March 2025

The following problem is from both the 2024 AMC 12A #5 and 2024 AMC 10A #7, so both problems redirect to this page.

Problem

Let $M$ be the midpoint of segment $\overline{AB}$ , and let $T$ lie on segment $\overline{AB}$ so that $AT \cdot AM = 100$ and $BT \cdot BM = 28$ . What is the length of segment $\overline{TM}$ ?

$\textbf{(A)}~4\qquad \textbf{(B)}~4.5\qquad \textbf{(C)}~5 \qquad \textbf{(D)}~5.5 \qquad \textbf{(E)}~6$

Solution

Note that $AM = BM$ , so $\tfrac{AT}{BT} = \tfrac{100}{28} = \tfrac{25}{7}$ . Thus, let $AT = 25x$ and $BT = 7x$ . Then $AM = BM = 16x$ and $TM = 9x$ . We can now use either of the two conditions presented, WLOG we use the first: $AT \cdot AM = (25x)(16x) = 400x^{2} = 100 \implies x = 0.5$ Therefore, $TM = 9x = \boxed{\textbf{(B)}~4.5}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 1: / Line 1: @@
+{{duplicate|[[2024 AMC 12A Problems/Problem 5|2024 AMC 12A #5]] and [[2024 AMC 10A Problems/Problem 7|2024 AMC 10A #7]]}}
 ==Problem==
+Let <math>M</math> be the midpoint of segment <math>\overline{AB}</math>, and let <math>T</math> lie on segment <math>\overline{AB}</math> so that <math>AT \cdot AM = 100</math> and <math>BT \cdot BM = 28</math>. What is the length of segment <math>\overline{TM}</math>?
-A data set containing <math>20</math> numbers, some of which are <math>6</math>, has mean <math>45</math>. When all the 6s are removed, the data set has mean <math>66</math>. How many 6s were in the original data set?
+<math>\textbf{(A)}~4\qquad \textbf{(B)}~4.5\qquad \textbf{(C)}~5 \qquad \textbf{(D)}~5.5 \qquad \textbf{(E)}~6</math>
-<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8</math>
-[[2024 AMC 12A Problems/Problem 5|Solution]]
-== Solution ==
-Because the set has <math>20</math> numbers and mean <math>45</math>, the sum of the terms in the set is <math>45\cdot 20=900</math>.
-Let there be <math>s</math> sixes in the set.
-Then, the mean of the set without the sixes is <math>\frac{900-6s}{20-s}</math>. Equating this expression to <math>66</math> and solving yields <math>s=7</math>, so we choose answer choice <math>\boxed{\textbf{(D) }7}</math>.
-==Solution 2==
-Let <math>S</math> be the sum of the data set without the sixes and <math>x</math> be the number of sixes. We are given that <math>\dfrac{S+6x}{20}=45</math> and <math>\dfrac S{20-x}=66</math>; the former equation becomes <math>S+6x=900</math> and the latter <math>S=1320-66x</math>. Since we want <math>x</math>, we equate the two equations and see that <math>900-6x=1320-66x\implies60x=420\implies x=\boxed{\textbf{(D) }7}</math>.
-~Technodoggo
-==Solution 3==
-Suppose there are <math>x</math> sixes. Then the sum of all the numbers can be written as <math>(20-x)\cdot 45+6x</math>
-Then, the mean of this set is <math>\frac{(20-x)\cdot 66+6x}{20}=45</math>. Solving this, we get <math>x=\boxed{\textbf{(D) }7}</math>
-== Video Solution by Math from my desk ==
-https://www.youtube.com/watch?v=E_Cab6NsbUA&t=2s
-== Video Solution 1 (⚡️ 1 min solve ⚡️) ==
-https://youtu.be/gGoqDf23XEk
-<i>~Education, the Study of Everything</i>
+==Solution==
+Note that <math>AM = BM</math>, so <math>\tfrac{AT}{BT} = \tfrac{100}{28} = \tfrac{25}{7}</math>. Thus, let <math>AT = 25x</math> and <math>BT = 7x</math>. Then <math>AM = BM = 16x</math> and <math>TM = 9x</math>. We can now use either of the two conditions presented, WLOG we use the first: <cmath>AT \cdot AM = (25x)(16x) = 400x^{2} = 100 \implies x = 0.5</cmath> Therefore, <math>TM = 9x = \boxed{\textbf{(B)}~4.5}</math>.
-== See Also ==
+==See also==
 {{AMC12 box|year=2024|ab=A|num-b=4|num-a=6}}
+{{AMC10 box|year=2024|ab=A|num-b=6|num-a=8}}
 {{MAA Notice}}

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Difference between revisions of "2024 AMC 12A Problems/Problem 5"

Revision as of 21:41, 20 March 2025

Problem

Solution

See also