Difference between revisions of "2021 AMC 12A Problems/Problem 17"
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− | {{duplicate|[[2021 AMC 10A Problems | + | {{duplicate|[[2021 AMC 10A Problems/Problem 17|2021 AMC 10A #17]] and [[2021 AMC 12A Problems/Problem 17|2021 AMC 12A #17]]}} |
==Problem== | ==Problem== | ||
Line 7: | Line 7: | ||
==Diagram== | ==Diagram== | ||
− | [[ | + | <asy> |
− | ~MRENTHUSIASM ( | + | /* Made by MRENTHUSIASM */ |
+ | size(300); | ||
+ | pair A, B, C, D, O, P; | ||
+ | C = (43,0); | ||
+ | D = (0,0); | ||
+ | B = intersectionpoints(Circle(C,43),Circle(D,66))[0]; | ||
+ | A = intersectionpoints(Circle(D,4*sqrt(190)),B--B+100*dir(180))[1]; | ||
+ | P = midpoint(B--D); | ||
+ | O = intersectionpoint(A--C,B--D); | ||
+ | dot("$C$",C,1.5*SE,linewidth(4)); | ||
+ | dot("$D$",D,1.5*SW,linewidth(4)); | ||
+ | dot("$B$",B,1.5*NE,linewidth(4)); | ||
+ | dot("$A$",A,1.5*NW,linewidth(4)); | ||
+ | dot("$P$",P,1.5*N,linewidth(4)); | ||
+ | dot("$O$",O,1.5*S,linewidth(4)); | ||
+ | markscalefactor=0.25; | ||
+ | draw(rightanglemark(A,D,O),red); | ||
+ | draw(A--B--C--D--cycle^^A--C^^B--D^^C--P); | ||
+ | label("$43$",B--C,E); | ||
+ | label("$43$",C--D,S); | ||
+ | label("$11$",midpoint(O--P),NW); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 1 (Similar Triangles and Pythagorean Theorem) == | ||
+ | Angle chasing* reveals that <math>\triangle BPC\sim\triangle BDA</math>, therefore | ||
+ | <cmath>2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},</cmath> | ||
+ | or <math>AB=86</math>. | ||
− | == | + | Additional angle chasing shows that <math>\triangle ABO\sim\triangle CDO</math>, therefore |
+ | <cmath>2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},</cmath> | ||
+ | or <math>BP=33</math> and <math>BD=66</math>. | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
Since <math>\triangle ADB</math> is right, the Pythagorean theorem implies that | Since <math>\triangle ADB</math> is right, the Pythagorean theorem implies that | ||
− | <cmath>AD=\sqrt{86^2-66^2} | + | <cmath>AD=\sqrt{86^2-66^2}=4\sqrt{190}.</cmath> |
− | + | The answer is <math>4+190=\boxed{\textbf{(D) }194}</math>. | |
− | <math> | + | |
+ | * Angle Chasing: If we set <math>\angle DBC = \alpha</math>, then we know that <math>\angle DCB = 180^\circ-2\alpha</math> because <math>\triangle DBC</math> is isosceles. Then, <math>\angle BCP = 90^\circ-\alpha</math>, so <math>\angle BPC</math> is a right angle. Because <math>\angle BDC = \alpha</math> and <math>\overline{AB}\parallel\overline{DC}</math>, we conclude that <math>\angle ABD = \alpha</math> too. Lastly, because <math>\triangle BPC</math> and <math>\triangle BDA</math> are both right triangles, they are similar by AA. | ||
− | ~mn28407 | + | ~mn28407 (Solution) |
+ | |||
+ | ~mm (Angle Chasing Remark) | ||
+ | |||
+ | ~eagleye ~MRENTHUSIASM ~charyyu83 (Minor Edits) | ||
==Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)== | ==Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)== | ||
− | Since <math>\triangle BCD</math> is isosceles with | + | Since <math>\triangle BCD</math> is isosceles with base <math>\overline{BD},</math> it follows that median <math>\overline{CP}</math> is also an altitude. Let <math>OD=x</math> and <math>CP=h,</math> so <math>PB=x+11.</math> |
− | Since <math>\angle AOD=\angle COP</math> by vertical angles, we | + | Since <math>\angle AOD=\angle COP</math> by vertical angles, we conclude that <math>\triangle AOD\sim\triangle COP</math> by AA, from which <math>\frac{AD}{CP}=\frac{OD}{OP},</math> or <cmath>AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.</cmath> Let the brackets denote areas. Notice that <math>[AOD]=[BOC]</math> (By the same base and height, we deduce that <math>[ACD]=[BDC].</math> Subtracting <math>[OCD]</math> from both sides gives <math>[AOD]=[BOC].</math>). Doubling both sides produces |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
2[AOD]&=2[BOC] \\ | 2[AOD]&=2[BOC] \\ | ||
Line 35: | Line 63: | ||
x^2&=11(x+22). | x^2&=11(x+22). | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Rearranging and factoring result <math>(x-22)(x+11)=0,</math> from which <math>x=22.</math> | + | Rearranging and factoring result in <math>(x-22)(x+11)=0,</math> from which <math>x=22.</math> |
− | Applying the Pythagorean Theorem to right <math>\triangle CPB,</math> we | + | Applying the Pythagorean Theorem to right <math>\triangle CPB,</math> we have <cmath>h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.</cmath> Finally, we get <cmath>AD=h\cdot\frac{x}{11}=4\sqrt{190},</cmath> so the answer is <math>4+190=\boxed{\textbf{(D) }194}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
==Solution 3 (Short)== | ==Solution 3 (Short)== | ||
− | Let <math>CP = y</math> | + | Let <math>CP = y</math>. <math>CP</math> a is perpendicular bisector of <math>DB.</math> Then, let <math>DO = x,</math> thus <math>DP = PB = 11+x.</math> |
(1) <math>\triangle CPO \sim \triangle ADO,</math> so we get <math>\frac{AD}{x} = \frac{y}{11},</math> or <math>AD = \frac{xy}{11}.</math> | (1) <math>\triangle CPO \sim \triangle ADO,</math> so we get <math>\frac{AD}{x} = \frac{y}{11},</math> or <math>AD = \frac{xy}{11}.</math> | ||
− | (2) | + | (2) Applying Pythagorean Theorem on <math>\triangle CDP</math> gives <math>(11+x)^2 + y^2 = 43^2.</math> |
− | (3) <math>\triangle BPC \sim \triangle BDA</math> with ratio <math>1:2,</math> so <math>AD = 2y | + | (3) <math>\triangle BPC \sim \triangle BDA</math> with ratio <math>1:2,</math> so <math>AD = 2y</math> using the fact that <math>P</math> is the midpoint of <math>BD</math>. |
− | Thus, <math>xy | + | Thus, <math>\frac{xy}{11} = 2y,</math> or <math>x = 22.</math> And <math>y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},</math> so <math>AD = 4 \sqrt{190}</math> and the answer is <math>4+190=\boxed{\textbf{(D) }194}.</math> |
~ ccx09 | ~ ccx09 | ||
− | ==Solution 4 | + | ==Solution 4 (Extending the Line)== |
Observe that <math>\triangle BPC</math> is congruent to <math>\triangle DPC</math>; both are similar to <math>\triangle BDA</math>. Let's extend <math>\overline{AD}</math> and <math>\overline{BC}</math> past points <math>D</math> and <math>C</math> respectively, such that they intersect at a point <math>E</math>. Observe that <math>\angle BDE</math> is <math>90</math> degrees, and that <math>\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA</math>. Thus, by ASA, we know that <math>\triangle ABD \cong \triangle EBD</math>, thus, <math>AD = ED</math>, meaning <math>D</math> is the midpoint of <math>AE</math>. | Observe that <math>\triangle BPC</math> is congruent to <math>\triangle DPC</math>; both are similar to <math>\triangle BDA</math>. Let's extend <math>\overline{AD}</math> and <math>\overline{BC}</math> past points <math>D</math> and <math>C</math> respectively, such that they intersect at a point <math>E</math>. Observe that <math>\angle BDE</math> is <math>90</math> degrees, and that <math>\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA</math>. Thus, by ASA, we know that <math>\triangle ABD \cong \triangle EBD</math>, thus, <math>AD = ED</math>, meaning <math>D</math> is the midpoint of <math>AE</math>. | ||
Let <math>M</math> be the midpoint of <math>\overline{DE}</math>. Note that <math>\triangle CME</math> is congruent to <math>\triangle BPC</math>, thus <math>BC = CE</math>, meaning <math>C</math> is the midpoint of <math>\overline{BE}.</math> | Let <math>M</math> be the midpoint of <math>\overline{DE}</math>. Note that <math>\triangle CME</math> is congruent to <math>\triangle BPC</math>, thus <math>BC = CE</math>, meaning <math>C</math> is the midpoint of <math>\overline{BE}.</math> | ||
Line 60: | Line 88: | ||
Therefore, <math>\overline{AC}</math> and <math>\overline{BD}</math> are both medians of <math>\triangle ABE</math>. This means that <math>O</math> is the centroid of <math>\triangle ABE</math>; therefore, because the centroid divides the median in a 2:1 ratio, <math>\frac{BO}{2} = DO = \frac{BD}{3}</math>. Recall that <math>P</math> is the midpoint of <math>BD</math>; <math>DP = \frac{BD}{2}</math>. The question tells us that <math>OP = 11</math>; <math>DP-DO=11</math>; we can write this in terms of <math>DB</math>; <math>\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66</math>. | Therefore, <math>\overline{AC}</math> and <math>\overline{BD}</math> are both medians of <math>\triangle ABE</math>. This means that <math>O</math> is the centroid of <math>\triangle ABE</math>; therefore, because the centroid divides the median in a 2:1 ratio, <math>\frac{BO}{2} = DO = \frac{BD}{3}</math>. Recall that <math>P</math> is the midpoint of <math>BD</math>; <math>DP = \frac{BD}{2}</math>. The question tells us that <math>OP = 11</math>; <math>DP-DO=11</math>; we can write this in terms of <math>DB</math>; <math>\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66</math>. | ||
+ | We are almost finished. Each side length of <math>\triangle ABD</math> is twice as long as the corresponding side length <math>\triangle CBP</math> or <math>\triangle CPD</math>, since those triangles are similar; this means that <math>AB = 2 \cdot 43 = 86</math>. Now, by Pythagorean theorem on <math>\triangle ABD</math>, <math>AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}</math>. | ||
− | + | The answer is <math>4+190 = \boxed{\textbf{(D) }194}</math>. | |
− | ~ ihatemath123 | + | ~ihatemath123 |
− | == Video Solution (Using Similar Triangles, Pythagorean Theorem) == | + | ==Solution 5== |
+ | Since <math>P</math> is the midpoint of isosceles triangle <math>BCD</math>, it would be pretty easy to see that <math>CP\perp BD</math>. Since <math>AD\perp BD</math> as well, <math>AD\parallel CP</math>. Connecting <math>AP</math>, it’s obvious that <math>[ADC]=[ADP]</math>. Since <math>DP=BP</math>, <math>[APB]=[ADC]</math>. | ||
+ | |||
+ | Since <math>P</math> is the midpoint of <math>BD</math>, the height of <math>\triangle APB</math> on side <math>AB</math> is half that of <math>\triangle ADC</math> on <math>CD</math>. Since <math>[APB]=[ADC]</math>, <math>AB=2CD</math>. | ||
+ | |||
+ | As a basic property of a trapezoid, <math>\triangle AOB \sim \triangle COD</math>, so <math>\frac{OB}{OD}=\frac{AB}{CD}=2</math>, or <math>OB=2OD</math>. Letting <math>OD=x</math>, then <math>PB=DP=11+x</math>, and <math>OB=22+x</math>. Hence <math>22+x=2x</math> and <math>x=22</math>. | ||
+ | |||
+ | Since <math>\triangle AOD \sim \triangle COP</math>, <math>\frac{AD}{PC}=\frac{OD}{OP}=2</math>. Since <math>PD=11+22=33</math>, <math>PC=\sqrt{43^2-33^2}=\sqrt{760}</math>. | ||
+ | |||
+ | So, <math>AD=2\sqrt{760}=4\sqrt{190}</math>. The correct answer is <math>\boxed{\textbf{(D) }194}</math>. | ||
+ | |||
+ | ==Solution 6 (Coordinate Geometry) == | ||
+ | |||
+ | Let <math>D</math> be the origin of the cartesian coordinate plane, <math>B</math> lie on the positive <math>x</math>-axis, and <math>A</math> lie on the negative <math>y</math>-axis. Then let the coordinates of <math>B = (2a,0), A = (0, -2b).</math> Then the slope of <math>AB</math> is <math>\frac{b}{a}.</math> Since <math>AB \parallel CD</math> the slope of <math>CD</math> is the same. Note that as <math>\triangle DCB</math> is isosceles <math>C</math> lies on <math>x = a.</math> Thus since <math>CD</math> has equation <math>y = \frac{b}{a}x</math> (<math>D</math> is the origin), <math>C = (a,b).</math> Therefore <math>AC</math> has equation <math>y = \frac{3b}{a}x - 2b</math> and intersects <math>BD</math> (<math>x</math>-axis) at <math>O =\left(\frac{2}{3}a, 0\right).</math> The midpoint of <math>BD</math> is <math>P = (a,0),</math> so <math>OP = \frac{a}{3} = 11,</math> from which <math>a = 33.</math> Then by Pythagorean theorem on <math>\triangle DPC</math> (<math>\triangle DBC</math> is isosceles), we have <math>b = \sqrt{43^2 - 33^2} = 2\sqrt{190},</math> so <math>2b=4\sqrt{190}.</math> | ||
+ | |||
+ | Finally, the answer is <math>4+190=\boxed{\textbf{(D) }194}.</math> | ||
+ | |||
+ | ~Aaryabhatta1 | ||
+ | |||
+ | ==Solution 7 (Trigonometry) == | ||
+ | |||
+ | [[Image:2021_12A_p17.png|thumb|center|600px|]] | ||
+ | |||
+ | set <math>\angle BDC = \theta </math> | ||
+ | |||
+ | BD = 2*DP = 2*43*Cos(<math>\theta</math>) | ||
+ | |||
+ | AB = BD / Cos(<math>\angle DBA</math>) = BD / Cos(<math>\theta</math>) = 2 *43 * Cos(<math>\theta</math>) / Cos(<math>\theta</math>) = 86 | ||
+ | |||
+ | OP/DO = CP / AD | ||
+ | |||
+ | 11 / (43Cos(<math>\theta</math>) - 11) = 43Sin(<math>\theta</math>) / 86 Sin(<math>\theta</math>) | ||
+ | |||
+ | Cos(<math>\theta</math>) = 33/ 43 | ||
+ | |||
+ | AD = 86 * Sin(<math>\theta</math>) = 2<math>\sqrt{760}= 4\sqrt{190}=\boxed{\textbf{(D) }194}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | == Video Solution by OmegaLearn (Using Similar Triangles, Pythagorean Theorem) == | ||
https://youtu.be/gjeSGJy_ld4 | https://youtu.be/gjeSGJy_ld4 | ||
Line 72: | Line 140: | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtube.com/watch?v=rtdovluzgQs | https://youtube.com/watch?v=rtdovluzgQs | ||
+ | |||
+ | == Video Solution by Mathematical Dexterity == | ||
+ | https://www.youtube.com/watch?v=QzAVdsgBBqg | ||
==See also== | ==See also== |
Latest revision as of 23:51, 23 July 2024
- The following problem is from both the 2021 AMC 10A #17 and 2021 AMC 12A #17, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Similar Triangles and Pythagorean Theorem)
- 4 Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)
- 5 Solution 3 (Short)
- 6 Solution 4 (Extending the Line)
- 7 Solution 5
- 8 Solution 6 (Coordinate Geometry)
- 9 Solution 7 (Trigonometry)
- 10 Video Solution by OmegaLearn (Using Similar Triangles, Pythagorean Theorem)
- 11 Video Solution by Punxsutawney Phil
- 12 Video Solution by Mathematical Dexterity
- 13 See also
Problem
Trapezoid has
, and
. Let
be the intersection of the diagonals
and
, and let
be the midpoint of
. Given that
, the length of
can be written in the form
, where
and
are positive integers and
is not divisible by the square of any prime. What is
?
Diagram
~MRENTHUSIASM
Solution 1 (Similar Triangles and Pythagorean Theorem)
Angle chasing* reveals that , therefore
or
.
Additional angle chasing shows that , therefore
or
and
.
Since is right, the Pythagorean theorem implies that
The answer is
.
- Angle Chasing: If we set
, then we know that
because
is isosceles. Then,
, so
is a right angle. Because
and
, we conclude that
too. Lastly, because
and
are both right triangles, they are similar by AA.
~mn28407 (Solution)
~mm (Angle Chasing Remark)
~eagleye ~MRENTHUSIASM ~charyyu83 (Minor Edits)
Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)
Since is isosceles with base
it follows that median
is also an altitude. Let
and
so
Since by vertical angles, we conclude that
by AA, from which
or
Let the brackets denote areas. Notice that
(By the same base and height, we deduce that
Subtracting
from both sides gives
). Doubling both sides produces
Rearranging and factoring result in
from which
Applying the Pythagorean Theorem to right we have
Finally, we get
so the answer is
~MRENTHUSIASM
Solution 3 (Short)
Let .
a is perpendicular bisector of
Then, let
thus
(1) so we get
or
(2) Applying Pythagorean Theorem on gives
(3) with ratio
so
using the fact that
is the midpoint of
.
Thus, or
And
so
and the answer is
~ ccx09
Solution 4 (Extending the Line)
Observe that is congruent to
; both are similar to
. Let's extend
and
past points
and
respectively, such that they intersect at a point
. Observe that
is
degrees, and that
. Thus, by ASA, we know that
, thus,
, meaning
is the midpoint of
.
Let
be the midpoint of
. Note that
is congruent to
, thus
, meaning
is the midpoint of
Therefore, and
are both medians of
. This means that
is the centroid of
; therefore, because the centroid divides the median in a 2:1 ratio,
. Recall that
is the midpoint of
;
. The question tells us that
;
; we can write this in terms of
;
.
We are almost finished. Each side length of is twice as long as the corresponding side length
or
, since those triangles are similar; this means that
. Now, by Pythagorean theorem on
,
.
The answer is .
~ihatemath123
Solution 5
Since is the midpoint of isosceles triangle
, it would be pretty easy to see that
. Since
as well,
. Connecting
, it’s obvious that
. Since
,
.
Since is the midpoint of
, the height of
on side
is half that of
on
. Since
,
.
As a basic property of a trapezoid, , so
, or
. Letting
, then
, and
. Hence
and
.
Since ,
. Since
,
.
So, . The correct answer is
.
Solution 6 (Coordinate Geometry)
Let be the origin of the cartesian coordinate plane,
lie on the positive
-axis, and
lie on the negative
-axis. Then let the coordinates of
Then the slope of
is
Since
the slope of
is the same. Note that as
is isosceles
lies on
Thus since
has equation
(
is the origin),
Therefore
has equation
and intersects
(
-axis) at
The midpoint of
is
so
from which
Then by Pythagorean theorem on
(
is isosceles), we have
so
Finally, the answer is
~Aaryabhatta1
Solution 7 (Trigonometry)
set
BD = 2*DP = 2*43*Cos()
AB = BD / Cos() = BD / Cos(
) = 2 *43 * Cos(
) / Cos(
) = 86
OP/DO = CP / AD
11 / (43Cos() - 11) = 43Sin(
) / 86 Sin(
)
Cos() = 33/ 43
AD = 86 * Sin() = 2
.
Video Solution by OmegaLearn (Using Similar Triangles, Pythagorean Theorem)
~ pi_is_3.14
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=rtdovluzgQs
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=QzAVdsgBBqg
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.