Difference between revisions of "2021 Fall AMC 10B Problems/Problem 15"
(added 'signature' to solution 2, (added solution 2 in previous edit, forgot to summarize it oops)) |
|||
Line 31: | Line 31: | ||
We have that <math>\triangle CRB \cong \triangle BAP.</math> Thus, <math>\frac{\overline{CB}}{\overline{CR}} = \frac{\overline{PB}}{\overline{AB}}</math>. Now, let the side length of the square be <math>s.</math> Then, by the Pythagorean theorem, <math>CR = \sqrt{x^2-36}.</math> Plugging all of this information in, we get <cmath>\frac{s}{\sqrt{s^2-36}} = \frac{13}{s}.</cmath> Simplifying gives <cmath>s^2=13\sqrt{s^2-36},</cmath> Squaring both sides gives <cmath>s^4 = 169s^2- 169\cdot 36 \implies s^4-169s^2 + 169\cdot 36 = 0.</cmath> We now set <math>s^2=t,</math> and get the equation <math>t^2-169t + 169\cdot 36 = 0.</math> From here, notice we want to solve for <math>t</math>, as it is precisely <math>s^2,</math> or the area of the square. So we use the [[Quadratic formula]], and though it may seem bashy, we hope for a nice cancellation of terms. <cmath>t = \frac{169\pm\sqrt{169^2-4\cdot 36 \cdot 169}}{2}.</cmath> It seems scary, but factoring <math>169</math> from the square root gives us <cmath>t = \frac{169\pm \sqrt{169 \cdot (169-144)}}{2} = \frac{169 \pm \sqrt{169 \cdot 25}}{2} = \frac{169 \pm 13\cdot 5}{2} = \frac{169\pm 65}{2},</cmath> giving us the solutions <cmath>t=52, 117.</cmath> We instantly see that <math>t=52</math> is way too small to be an area of this square (<math>52</math> isn't even an answer choice, so you can skip this step if out of time) because then the side length would be <math>2\sqrt{13}</math> and then, even the largest line you can draw inside the square (the diagonal) is <math>2\sqrt{26},</math> which is less than <math>13</math> (line <math>PB</math>) And thus, <math>t</math> must be <math>117</math>, and our answer is <math>\boxed{\textbf{(D)}}.</math> <math>\blacksquare</math> | We have that <math>\triangle CRB \cong \triangle BAP.</math> Thus, <math>\frac{\overline{CB}}{\overline{CR}} = \frac{\overline{PB}}{\overline{AB}}</math>. Now, let the side length of the square be <math>s.</math> Then, by the Pythagorean theorem, <math>CR = \sqrt{x^2-36}.</math> Plugging all of this information in, we get <cmath>\frac{s}{\sqrt{s^2-36}} = \frac{13}{s}.</cmath> Simplifying gives <cmath>s^2=13\sqrt{s^2-36},</cmath> Squaring both sides gives <cmath>s^4 = 169s^2- 169\cdot 36 \implies s^4-169s^2 + 169\cdot 36 = 0.</cmath> We now set <math>s^2=t,</math> and get the equation <math>t^2-169t + 169\cdot 36 = 0.</math> From here, notice we want to solve for <math>t</math>, as it is precisely <math>s^2,</math> or the area of the square. So we use the [[Quadratic formula]], and though it may seem bashy, we hope for a nice cancellation of terms. <cmath>t = \frac{169\pm\sqrt{169^2-4\cdot 36 \cdot 169}}{2}.</cmath> It seems scary, but factoring <math>169</math> from the square root gives us <cmath>t = \frac{169\pm \sqrt{169 \cdot (169-144)}}{2} = \frac{169 \pm \sqrt{169 \cdot 25}}{2} = \frac{169 \pm 13\cdot 5}{2} = \frac{169\pm 65}{2},</cmath> giving us the solutions <cmath>t=52, 117.</cmath> We instantly see that <math>t=52</math> is way too small to be an area of this square (<math>52</math> isn't even an answer choice, so you can skip this step if out of time) because then the side length would be <math>2\sqrt{13}</math> and then, even the largest line you can draw inside the square (the diagonal) is <math>2\sqrt{26},</math> which is less than <math>13</math> (line <math>PB</math>) And thus, <math>t</math> must be <math>117</math>, and our answer is <math>\boxed{\textbf{(D)}}.</math> <math>\blacksquare</math> | ||
+ | |||
+ | ~wamofan | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=16|num-b=14}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=16|num-b=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:54, 24 November 2021
Contents
[hide]Problem
In square , points
and
lie on
and
, respectively. Segments
and
intersect at right angles at
, with
and
. What is the area of the square?
Solution
Note that Then, it follows that
Thus,
Define
to be the length of side
then
Because
is the altitude of the triangle, we can use the property that
Substituting the given lengths, we have
Solving, gives
and
We eliminate the possibilty of
because
Thus, the side lengnth of the square, by Pythagorean Theorem, is
Thus, the area of the sqaure is
Thus, the answer is
~NH14
Solution 2
We have that Thus,
. Now, let the side length of the square be
Then, by the Pythagorean theorem,
Plugging all of this information in, we get
Simplifying gives
Squaring both sides gives
We now set
and get the equation
From here, notice we want to solve for
, as it is precisely
or the area of the square. So we use the Quadratic formula, and though it may seem bashy, we hope for a nice cancellation of terms.
It seems scary, but factoring
from the square root gives us
giving us the solutions
We instantly see that
is way too small to be an area of this square (
isn't even an answer choice, so you can skip this step if out of time) because then the side length would be
and then, even the largest line you can draw inside the square (the diagonal) is
which is less than
(line
) And thus,
must be
, and our answer is
~wamofan
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.