Difference between revisions of "2021 Fall AMC 10B Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
− | Note that <math>\triangle APB \cong \triangle BQC.</math> Then, it follows that <math>\overline{PB} \cong \overline{QC}.</math> Thus, <math>QC = PB = PR + RB = 7 + 6 = 13.</math> Define <math>x</math> to be the length of side <math>CR,</math> then <math>RQ = 13-x.</math> Because <math>\overline{BR}</math> is the altitude of the triangle, we can use the property that <math>QR \cdot RC = BR^2.</math> Substituting the given lengths, we have <cmath>(13-x) \cdot x = 36.</cmath> Solving, gives <math> | + | Note that <math>\triangle APB \cong \triangle BQC</math> by ASA. (<math>\angle PAB = \angle QBC = 90^\circ, AB=CB,</math> and <math>\angle PBA = \angle QCB.</math>) Then, it follows that <math>\overline{PB} \cong \overline{QC}.</math> Thus, <math>QC = PB = PR + RB = 7 + 6 = 13.</math> Define <math>x</math> to be the length of side <math>CR,</math> then <math>RQ = 13-x.</math> Because <math>\overline{BR}</math> is the altitude of the triangle, we can use the property that <math>QR \cdot RC = BR^2.</math> Substituting the given lengths, we have <cmath>(13-x) \cdot x = 36.</cmath> Solving, gives <math>RQ = 4</math> and <math>RC = 9.</math> We eliminate the possibility of <math>x=4</math> because <math>RC > QR.</math> Thus, the side length of the square, by Pythagorean Theorem, is <cmath>\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.</cmath> Thus, the area of the square is <math>(\sqrt{117})^2 = 117,</math> so the answer is <math>\boxed{\textbf{(D) }117}.</math> |
− | + | Note that there is another way to prove that <math>CR = 4</math> is impossible. If <math>CR = 4,</math> then the side length would be <math>\sqrt{4^2 + 6^2} = \sqrt{52},</math> and the area would be <math>52,</math> but that isn't in the answer choices. Thus, <math>CR</math> must be <math>9.</math> | |
− | + | ~NH14 ~sl_hc | |
− | + | Extra Note: Another way to prove <math>4</math> is impossible. The side length of the square, <math>S</math>, is equal to <math>\sqrt{4^2 + 6^2} = \sqrt{52}</math>. Because <math>x = 4</math>, <math>RQ = 9</math>. Because <math>QB = \sqrt{RB^2 + RQ^2} = \sqrt{6^2 + 9^2} = \sqrt{117}</math> and <math>QB < S</math> but <math>\sqrt{117} > \sqrt{52}</math>, we have proof by contradiction. And so <math>x = 9</math>. | |
− | + | ~ Wiselion (Extra Note) | |
− | |||
− | We want <math>x^2</math>, so we want to find <math>x</math>. | + | ==Solution 2 (Similarity, Pythagorean Theorem, and Systems of Equations)== |
+ | |||
+ | As above, note that <math>\bigtriangleup BPA \cong \bigtriangleup CQB</math>, which means that <math>QC = 13</math>. In addition, note that <math>BR</math> is the altitude of a right triangle to its hypotenuse, so <math>\bigtriangleup BQR \sim \bigtriangleup CBR \sim \bigtriangleup CQB</math>. Let the side length of the square be <math>x</math>; using similarity side ratios of <math>\bigtriangleup BQR</math> to <math>\bigtriangleup CQB</math>, we get <cmath>\frac{6}{x} = \frac{QB}{13} \implies QB \cdot x = 78</cmath>Note that <math>QB^2 + x^2 = 13^2 = 169</math> by the Pythagorean theorem, so we can use the expansion <math>(a+b)^2 = a^2+2ab+b^2</math> to produce two equations and two variables; | ||
+ | |||
+ | <cmath>(QB + x)^2 = QB^2 + 2QB\cdot x + x^2 \implies (QB+x)^2 = 169 + 2 \cdot 78 \implies QB+x = \sqrt{13(13)+13(12)} = \sqrt{13 \cdot 25} = 5\sqrt{13}</cmath> | ||
+ | <cmath>(QB-x)^2 = QB^2 - 2QB\cdot x + x^2 \implies (QB - x)^2 = 169 - 2\cdot 78 \implies \pm(QB-x) = \sqrt{13(13) - 13(12)}</cmath> | ||
+ | |||
+ | Since <math>QB-x</math> is negative, it doesn't make sense in the context of this problem, so we go with <cmath>x-QB = \sqrt{13(13) - 13(12)} = \sqrt{13 \cdot 1} = \sqrt{13}</cmath> | ||
+ | |||
+ | We want <math>x^2</math>, so we want to find <math>x</math>. Adding the first equation to the second, we get <cmath>2x = 6\sqrt{13} \implies x = 3\sqrt{13}</cmath> | ||
Then <math>x^2</math> = <math>(3\sqrt{13}^2) = 9 \cdot 13 = 117 = \boxed{D}</math> | Then <math>x^2</math> = <math>(3\sqrt{13}^2) = 9 \cdot 13 = 117 = \boxed{D}</math> | ||
+ | |||
~KingRavi | ~KingRavi | ||
+ | |||
+ | ~stjwyl (Edits) | ||
+ | |||
+ | -yingkai_0_ (Minor Edits) | ||
==Solution 3== | ==Solution 3== | ||
Line 63: | Line 76: | ||
which reduces to <math>|13a| = a^2 + 36</math>. Since <math>a</math> is positive, the last equations factors as | which reduces to <math>|13a| = a^2 + 36</math>. Since <math>a</math> is positive, the last equations factors as | ||
<math> a^2 - 13a + 36 = (a-4)(a-9) = 0 </math>. Now judging from the figure, we learn that <math>a > RB = 6</math>. So <math>a = 9</math>. | <math> a^2 - 13a + 36 = (a-4)(a-9) = 0 </math>. Now judging from the figure, we learn that <math>a > RB = 6</math>. So <math>a = 9</math>. | ||
− | Therefore, the area of the square <math>ABCD</math> is <math>BC^2 = RC^2 + RB^2 = a^2 + 6^2 = | + | Therefore, the area of the square <math>ABCD</math> is <math>BC^2 = RC^2 + RB^2 = a^2 + 6^2 = \boxed{\textbf{(D)}\ 117}</math>. |
~VensL. | ~VensL. | ||
+ | can someone explain how line bc doesnt have an undefined slope?? how are there 2 different x coordinates if its a vertical line... | ||
+ | ~THATONEKID_-TOKYT- | ||
+ | The line is not vertical. Looking closely, you can actually see that CQ and PB lie on the Y and X axis respectively. | ||
+ | |||
+ | == Solution 5 == | ||
+ | Denote <math>\angle PBA = \alpha</math>. | ||
+ | Because <math>\angle QRB = \angle QBC = 90^\circ</math>, <math>\angle BCQ = \alpha</math>. | ||
+ | |||
+ | Hence, <math>AB = BP \cos \angle PBA = 13 \cos \alpha</math>, <math>BC = \frac{BR}{\sin \angle BCQ} = \frac{6}{\sin \alpha}</math>. | ||
+ | |||
+ | Because <math>ABCD</math> is a square, <math>AB = BC</math>. | ||
+ | Hence, <math>13 \cos \alpha = \frac{6}{\sin \alpha}</math>. | ||
+ | |||
+ | Therefore, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sin 2 \alpha & = 2 \sin \alpha \cos \alpha \\ | ||
+ | & = \frac{12}{13} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Thus, <math>\cos 2 \alpha = \pm \frac{5}{13}</math>. | ||
+ | |||
+ | <math>\textbf{Case 1}</math>: <math>\cos 2 \alpha = \frac{5}{13}</math>. | ||
+ | |||
+ | Thus, <math>\cos \alpha = \sqrt{\frac{1 + \cos 2 \alpha}{2}} = \frac{3}{\sqrt{13}}</math>. | ||
+ | |||
+ | Hence, <math>AB = 13 \cos \alpha = 3 \sqrt{13}</math>. | ||
+ | |||
+ | Therefore, <math>{\rm Area} \ ABCD = AB^2 = 117</math>. | ||
+ | |||
+ | <math>\textbf{Case 2}</math>: <math>\cos 2 \alpha = - \frac{5}{13}</math>. | ||
+ | |||
+ | Thus, <math>\cos \alpha = \sqrt{\frac{1 + \cos 2 \alpha}{2}} = \frac{2}{\sqrt{13}}</math>. | ||
+ | |||
+ | Hence, <math>AB = 13 \cos \alpha = 2 \sqrt{13}</math>. | ||
+ | |||
+ | However, we observe <math>BQ = \frac{BR}{\cos \alpha} = 3 \sqrt{13} > AB</math>. | ||
+ | Therefore, in this case, point <math>Q</math> is not on the segment <math>AB</math>. | ||
+ | |||
+ | Therefore, this case is infeasible. | ||
+ | |||
+ | Putting all cases together, the answer is <math>\boxed{\textbf{(D) }117}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 6 (Answer choices and areas)== | ||
+ | Note that if we connect points <math>P</math> and <math>C</math>, we get a triangle with height <math>RC</math> and length <math>13</math>. This triangle has an area of <math>\frac {1}{2}</math> the square. We can now use answer choices to our advantage! | ||
+ | |||
+ | Answer choice A: If <math>BC</math> was <math>\sqrt {85}</math>, <math>RC</math> would be <math>7</math>. The triangle would therefore have an area of <math>\frac {91}{2}</math> which is not half of the area of the square. Therefore, A is wrong. | ||
+ | |||
+ | Answer choice B: If <math>BC</math> was <math>\sqrt {93}</math>, <math>RC</math> would be <math>\sqrt {57}</math>. This is obviously wrong. | ||
+ | |||
+ | Answer choice C: If <math>BC</math> was <math>10</math>, we would have that <math>RC</math> is <math>8</math>. The area of the triangle would be <math>52</math>, which is not half the area of the square. Therefore, C is wrong. | ||
+ | |||
+ | Answer choice D: If <math>BC</math> was <math>\sqrt {117}</math>, that would mean that <math>RC</math> is <math>9</math>. The area of the triangle would therefore be <math>\frac {117}{2}</math> which IS half the area of the square. Therefore, our answer is <math>\boxed {\textbf{(D) 117}}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | ==Solution 7 (Power of a Point)== | ||
+ | Note that <math>PRQA</math> is a cyclic quadrilateral (opposite angles add to <math>180^{\circ}</math>). Call the circumcircle of quadrilateral <math>PRQA</math> <math>O</math>. Then the power of <math>B</math> to <math>O</math> is <math>6\cdot (6+7)=78</math>. Let <math>a</math> be the length of <math>BQ</math> and <math>s</math> the side length of the square, then we have <math>a\cdot s = 78</math>, and we also have <math>a^2 + s^2=13^2</math>, solving the two equation will give us <math>s^2=117</math>. | ||
+ | |||
+ | ~student99 | ||
+ | |||
+ | ~minor edits by [[User: Yiyj1|Yiyj1]] | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://www.youtube.com/watch?v=sKC0Yt6sPi0 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/_6o7d9pGJng | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/p9Hq6N-cEAM | ||
+ | |||
+ | ~savannahsolver | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/R7TwXgAGYuw?t=1367 (note in the comments an easier solution too from a viewer) | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/hDsoyvFWYxc?t=822 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==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}} |
Latest revision as of 17:33, 6 June 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Similarity, Pythagorean Theorem, and Systems of Equations)
- 4 Solution 3
- 5 Solution 4 (Point-line distance formula)
- 6 Solution 5
- 7 Solution 6 (Answer choices and areas)
- 8 Solution 7 (Power of a Point)
- 9 Video Solution by Interstigation
- 10 Video Solution
- 11 Video Solution by WhyMath
- 12 Video Solution by TheBeautyofMath
- 13 Video Solution by OmegaLearn
- 14 See Also
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 1
Note that by ASA. (
and
) 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 possibility of
because
Thus, the side length of the square, by Pythagorean Theorem, is
Thus, the area of the square is
so the answer is
Note that there is another way to prove that is impossible. If
then the side length would be
and the area would be
but that isn't in the answer choices. Thus,
must be
~NH14 ~sl_hc
Extra Note: Another way to prove is impossible. The side length of the square,
, is equal to
. Because
,
. Because
and
but
, we have proof by contradiction. And so
.
~ Wiselion (Extra Note)
Solution 2 (Similarity, Pythagorean Theorem, and Systems of Equations)
As above, note that , which means that
. In addition, note that
is the altitude of a right triangle to its hypotenuse, so
. Let the side length of the square be
; using similarity side ratios of
to
, we get
Note that
by the Pythagorean theorem, so we can use the expansion
to produce two equations and two variables;
Since is negative, it doesn't make sense in the context of this problem, so we go with
We want , so we want to find
. Adding the first equation to the second, we get
Then =
~KingRavi
~stjwyl (Edits)
-yingkai_0_ (Minor Edits)
Solution 3
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
Solution 4 (Point-line distance formula)
Denote . Now tilt your head to the right and view
and
as the origin,
-axis and
-axis, respectively. In particular, we have points
. Note that side length of the square
is
. Also equation of line
is
Because the distance from
to line
is also the side length
, we can apply the point-line distance formula to get
which reduces to
. Since
is positive, the last equations factors as
. Now judging from the figure, we learn that
. So
.
Therefore, the area of the square
is
.
~VensL. can someone explain how line bc doesnt have an undefined slope?? how are there 2 different x coordinates if its a vertical line... ~THATONEKID_-TOKYT- The line is not vertical. Looking closely, you can actually see that CQ and PB lie on the Y and X axis respectively.
Solution 5
Denote .
Because
,
.
Hence, ,
.
Because is a square,
.
Hence,
.
Therefore,
Thus, .
:
.
Thus, .
Hence, .
Therefore, .
:
.
Thus, .
Hence, .
However, we observe .
Therefore, in this case, point
is not on the segment
.
Therefore, this case is infeasible.
Putting all cases together, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 6 (Answer choices and areas)
Note that if we connect points and
, we get a triangle with height
and length
. This triangle has an area of
the square. We can now use answer choices to our advantage!
Answer choice A: If was
,
would be
. The triangle would therefore have an area of
which is not half of the area of the square. Therefore, A is wrong.
Answer choice B: If was
,
would be
. This is obviously wrong.
Answer choice C: If was
, we would have that
is
. The area of the triangle would be
, which is not half the area of the square. Therefore, C is wrong.
Answer choice D: If was
, that would mean that
is
. The area of the triangle would therefore be
which IS half the area of the square. Therefore, our answer is
.
~Arcticturn
Solution 7 (Power of a Point)
Note that is a cyclic quadrilateral (opposite angles add to
). Call the circumcircle of quadrilateral
. Then the power of
to
is
. Let
be the length of
and
the side length of the square, then we have
, and we also have
, solving the two equation will give us
.
~student99
~minor edits by Yiyj1
Video Solution by Interstigation
https://www.youtube.com/watch?v=sKC0Yt6sPi0
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/R7TwXgAGYuw?t=1367 (note in the comments an easier solution too from a viewer)
~IceMatrix
Video Solution by OmegaLearn
https://youtu.be/hDsoyvFWYxc?t=822
~ pi_is_3.14
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.