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Difference between revisions of "2009 AMC 10B Problems/Problem 10"

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(Note that the resulting triangle is the well-known <math>5-12-13</math> right triangle, scaled by <math>1/5</math>.)
 
(Note that the resulting triangle is the well-known <math>5-12-13</math> right triangle, scaled by <math>1/5</math>.)
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== Solution 2 ==
 
== Solution 2 ==
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== Solution 3 ==
 
== Solution 3 ==
  
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<asy>
 
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size(300);
 
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pair A, B, C, D, E, x;
 
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A =(0, 5);
 
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B = (0, 0);
 
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C = (3, 0);
 
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D = (0, 1.6);
 
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E = (A+C)/2;
 
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x = (B+D)/2;
 
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draw(MP("A",A, (0, 1))--MP(P("C",C,(1, -1))--cycle);
 
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draw(C--MP("D",D, (-1, 0))--MP("E",E,(1, 1)));
 
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MP("x",x, (-1, 0));
 
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draw(rightanglemark(A, B, C));
 
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draw(rightanglemark(A, E, D));
 
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draw(anglemark(B, A, C));
 
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</asy>
 
  
 
Let <math>AB</math> represent the flagpole in the diagram above. After the flagpole breaks at point <math>D</math>, its tip lies at point <math>C</math>. Since none of the flagpole is destroyed, we know that <math>DA=DC</math>. Therefore, triangle <math>\triangle ADC</math> is isosceles.
 
Let <math>AB</math> represent the flagpole in the diagram above. After the flagpole breaks at point <math>D</math>, its tip lies at point <math>C</math>. Since none of the flagpole is destroyed, we know that <math>DA=DC</math>. Therefore, triangle <math>\triangle ADC</math> is isosceles.
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Since <math>AB = 5</math> and <math>BC = 1</math>, we have that <math>AD = \frac{5^2 + 1^2}{2 \cdot 5} = 2.6</math>, and thus <math>x = AB - AD = \boxed{2.4}</math>.
 
Since <math>AB = 5</math> and <math>BC = 1</math>, we have that <math>AD = \frac{5^2 + 1^2}{2 \cdot 5} = 2.6</math>, and thus <math>x = AB - AD = \boxed{2.4}</math>.
 +
 +
==Video Solution==
 +
 +
There is currently no video solution on Youtube. A wrong link was posted here.
 +
 +
(I sent a request to BeautyofMath so there may be one on his channel soon)
 +
 +
 +
~ ssr-07
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:12, 12 July 2023

Problem

A flagpole is originally $5$ meters tall. A hurricane snaps the flagpole at a point $x$ meters above the ground so that the upper part, still attached to the stump, touches the ground $1$ meter away from the base. What is $x$?

$\text{(A) } 2.0 \qquad \text{(B) } 2.1 \qquad \text{(C) } 2.2 \qquad \text{(D) } 2.3 \qquad \text{(E) } 2.4$

Solution 1

The broken flagpole forms a right triangle with legs $1$ and $x$, and hypotenuse $5-x$. The Pythagorean theorem now states that $1^2 + x^2 = (5-x)^2$, hence $10x = 24$, and $x=\boxed{2.4}$.

(Note that the resulting triangle is the well-known $5-12-13$ right triangle, scaled by $1/5$.)

Solution 2

A right triangle is formed with the bottom of the flagpole, the snapped part, and the ground. One leg is of length $1$ and the other is length $x$. By the Pythagorean theorem, we know that $\sqrt{x^2+1^2}$ must be the length of the snapped part of the flagpole. Observe that all the answer choices are rational. If $x$ is rational, $5-x$, which is the snapped part, must also be rational. Therefore, $1, x, 5-x$ must form a scaled Pythagorean triple. We know that $10, 24, 26$ is a Pythagorean triple, so the corresponding answer must be $1, 2.4, 2.6$. Adding together the $x$ and the snapped part, this does indeed equal $5$, so our solution is done.

Solution 3

Let $AB$ represent the flagpole in the diagram above. After the flagpole breaks at point $D$, its tip lies at point $C$. Since none of the flagpole is destroyed, we know that $DA=DC$. Therefore, triangle $\triangle ADC$ is isosceles.

Draw the altitude $DE \perp AC$. Since $\triangle ADC$ is isosceles, we know that $AE = EC$. Also note that $\triangle AED \sim \triangle ABC$. Therefore, \begin{align*} AD &= AE \times \frac{AD}{AE} \\ &= \frac{AC}{2} \times \frac{AC}{AB} \\ &= \frac{AC^2}{2 AB} \\ &= \frac{AB^2 + BC^2}{2 AB} \end{align*}

Since $AB = 5$ and $BC = 1$, we have that $AD = \frac{5^2 + 1^2}{2 \cdot 5} = 2.6$, and thus $x = AB - AD = \boxed{2.4}$.

Video Solution

There is currently no video solution on Youtube. A wrong link was posted here.

(I sent a request to BeautyofMath so there may be one on his channel soon)


~ ssr-07

See Also

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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

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