Difference between revisions of "2023 AIME I Problems/Problem 8"
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==Diagram== | ==Diagram== | ||
<asy> | <asy> | ||
− | /* Made by MRENTHUSIASM */ | + | /* Made by MRENTHUSIASM; inspired by Math Jams. */ |
size(300); | size(300); | ||
Line 20: | Line 20: | ||
markscalefactor=0.15; | markscalefactor=0.15; | ||
draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); | draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); | ||
− | draw(Circle(O,25/2)^^A--B--C--D--cycle | + | draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T); |
− | + | draw(P--R^^P--S^^P--T,red+dashed); | |
− | draw(P--R^^P--S^^P--T,red); | ||
dot("$A$",A,1.5*dir(225),linewidth(4.5)); | dot("$A$",A,1.5*dir(225),linewidth(4.5)); | ||
dot("$B$",B,1.5*dir(-45),linewidth(4.5)); | dot("$B$",B,1.5*dir(-45),linewidth(4.5)); | ||
Line 40: | Line 39: | ||
This solution refers to the <b>Diagram</b> section. | This solution refers to the <b>Diagram</b> section. | ||
+ | |||
+ | Let <math>O</math> be the incenter of <math>ABCD</math> for which <math>\odot O</math> is tangent to <math>\overline{DA},\overline{AB},</math> and <math>\overline{BC}</math> at <math>X,Y,</math> and <math>Z,</math> respectively. Moreover, suppose that <math>R,S,</math> and <math>T</math> are the feet of the perpendiculars from <math>P</math> to <math>\overleftrightarrow{DA},\overleftrightarrow{AB},</math> and <math>\overleftrightarrow{BC},</math> respectively, such that <math>\overline{RT}</math> intersects <math>\odot O</math> at <math>P</math> and <math>Q.</math> | ||
+ | |||
+ | We obtain the following diagram: | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM; inspired by Math Jams. */ | ||
+ | |||
+ | size(300); | ||
+ | pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q; | ||
+ | A = origin; | ||
+ | B = (125/4,0); | ||
+ | C = B + 125/4 * dir((3,4)); | ||
+ | D = A + 125/4 * dir((3,4)); | ||
+ | O = (25,25/2); | ||
+ | P = (15,5); | ||
+ | R = foot(P,A,D); | ||
+ | S = foot(P,A,B); | ||
+ | T = foot(P,B,C); | ||
+ | X = (15,20); | ||
+ | Y = (25,0); | ||
+ | Z = (35,5); | ||
+ | Q = intersectionpoints(Circle(O,25/2),R--T)[1]; | ||
+ | |||
+ | fill(R--T--Z--X--cycle,cyan); | ||
+ | markscalefactor=0.15; | ||
+ | draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); | ||
+ | draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T); | ||
+ | draw(P--R^^P--S^^P--T,red+dashed); | ||
+ | draw(O--X^^O--Y^^O--Z); | ||
+ | dot("$A$",A,1.5*dir(225),linewidth(4.5)); | ||
+ | dot("$B$",B,1.5*dir(-45),linewidth(4.5)); | ||
+ | dot("$C$",C,1.5*dir(45),linewidth(4.5)); | ||
+ | dot("$D$",D,1.5*dir(135),linewidth(4.5)); | ||
+ | dot("$P$",P,1.5*dir(60),linewidth(4.5)); | ||
+ | dot("$R$",R,1.5*dir(135),linewidth(4.5)); | ||
+ | dot("$S$",S,1.5*dir(-90),linewidth(4.5)); | ||
+ | dot("$T$",T,1.5*dir(-45),linewidth(4.5)); | ||
+ | dot("$O$",O,1.5*dir(45),linewidth(4.5)); | ||
+ | dot("$X$",X,1.5*dir(135),linewidth(4.5)); | ||
+ | dot("$Y$",Y,1.5*dir(-90),linewidth(4.5)); | ||
+ | dot("$Z$",Z,1.5*dir(-45),linewidth(4.5)); | ||
+ | dot("$Q$",Q,1.5*dir(60),linewidth(4.5)); | ||
+ | |||
+ | label("$9$",midpoint(P--R),dir(A-D),red); | ||
+ | label("$5$",midpoint(P--S),dir(180),red); | ||
+ | label("$16$",midpoint(P--T),dir(A-D),red); | ||
+ | </asy> | ||
+ | Note that <math>\angle RXZ = \angle TZX = 90^\circ</math> by the properties of tangents, so <math>RTZX</math> is a rectangle. It follows that the diameter of <math>\odot O</math> is <math>XZ = RT = 25.</math> | ||
+ | |||
+ | Let <math>x=PQ</math> and <math>y=RX=TZ.</math> | ||
+ | We apply the Power of a Point Theorem to <math>R</math> and <math>T:</math> | ||
+ | <cmath>\begin{align*} | ||
+ | y^2 &= 9(9+x), \\ | ||
+ | y^2 &= 16(16-x). | ||
+ | \end{align*}</cmath> | ||
+ | We solve this system of equations to get <math>x=7</math> and <math>y=12.</math> Alternatively, we can find these results by the symmetry on rectangle <math>RTZX</math> and semicircle <math>\widehat{XPZ}.</math> | ||
+ | |||
+ | We extend <math>\overline{SP}</math> beyond <math>P</math> to intersect <math>\odot O</math> and <math>\overleftrightarrow{CD}</math> at <math>E</math> and <math>F,</math> respectively, where <math>E\neq P.</math> So, we have <math>EF=SP=5</math> and <math>PE=25-SP-EF=15.</math> On the other hand, we have <math>PX=15</math> by the Pythagorean Theorem on right <math>\triangle PRX.</math> Together, we conclude that <math>E=X.</math> Therefore, points <math>S,P,</math> and <math>X</math> must be collinear. | ||
+ | |||
+ | Let <math>G</math> be the foot of the perpendicular from <math>D</math> to <math>\overline{AB}.</math> Note that <math>\overline{DG}\parallel\overline{XP},</math> as shown below: | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM; inspired by Math Jams. */ | ||
+ | |||
+ | size(300); | ||
+ | pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q, G; | ||
+ | A = origin; | ||
+ | B = (125/4,0); | ||
+ | C = B + 125/4 * dir((3,4)); | ||
+ | D = A + 125/4 * dir((3,4)); | ||
+ | O = (25,25/2); | ||
+ | P = (15,5); | ||
+ | R = foot(P,A,D); | ||
+ | S = foot(P,A,B); | ||
+ | T = foot(P,B,C); | ||
+ | X = (15,20); | ||
+ | Y = (25,0); | ||
+ | Z = (35,5); | ||
+ | Q = intersectionpoints(Circle(O,25/2),R--T)[1]; | ||
+ | G = foot(D,A,B); | ||
+ | |||
+ | fill(D--A--G--cycle,green); | ||
+ | fill(P--R--X--cycle,yellow); | ||
+ | markscalefactor=0.15; | ||
+ | draw(rightanglemark(P,R,D)^^rightanglemark(D,G,A),red); | ||
+ | draw(Circle(O,25/2)^^A--B--C--D--cycle^^X--P^^D--G); | ||
+ | draw(P--R,red+dashed); | ||
+ | dot("$A$",A,1.5*dir(225),linewidth(4.5)); | ||
+ | dot("$B$",B,1.5*dir(-45),linewidth(4.5)); | ||
+ | dot("$C$",C,1.5*dir(45),linewidth(4.5)); | ||
+ | dot("$D$",D,1.5*dir(135),linewidth(4.5)); | ||
+ | dot("$P$",P,1.5*dir(60),linewidth(4.5)); | ||
+ | dot("$R$",R,1.5*dir(135),linewidth(4.5)); | ||
+ | dot("$O$",O,1.5*dir(45),linewidth(4.5)); | ||
+ | dot("$X$",X,1.5*dir(135),linewidth(4.5)); | ||
+ | dot("$G$",G,1.5*dir(-90),linewidth(4.5)); | ||
+ | draw(P--X,MidArrow(0.3cm,Fill(red))); | ||
+ | draw(G--D,MidArrow(0.3cm,Fill(red))); | ||
+ | |||
+ | label("$9$",midpoint(P--R),dir(A-D),red); | ||
+ | label("$12$",midpoint(R--X),dir(135),red); | ||
+ | label("$15$",midpoint(X--P),dir(0),red); | ||
+ | label("$25$",midpoint(G--D),dir(0),red); | ||
+ | </asy> | ||
+ | As <math>\angle PRX = \angle AGD = 90^\circ</math> and <math>\angle PXR = \angle ADG</math> by the AA Similarity, we conclude that <math>\triangle PRX \sim \triangle AGD.</math> The ratio of similitude is <cmath>\frac{PX}{AD} = \frac{RX}{GD}.</cmath> We get <math>\frac{15}{AD} = \frac{12}{25},</math> from which <math>AD = \frac{125}{4}.</math> | ||
+ | |||
+ | Finally, the perimeter of <math>ABCD</math> is <math>4AD = \boxed{125}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM (inspired by awesomeming327. and WestSuburb) | ||
==Solution 2== | ==Solution 2== | ||
+ | |||
+ | This solution refers to the <b>Diagram</b> section. | ||
+ | |||
+ | Define points <math>O,R,S,</math> and <math>T</math> as Solution 1 does. Moreover, let | ||
+ | <math>H</math> be the foot of the perpendicular from <math>P</math> to <math>\overleftrightarrow{CD},</math> | ||
+ | <math>M</math> be the foot of the perpendicular from <math>O</math> to <math>\overleftrightarrow{HS},</math> and | ||
+ | <math>N</math> be the foot of the perpendicular from <math>O</math> to <math>\overleftrightarrow{RT}.</math> | ||
+ | |||
+ | We obtain the following diagram: | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM; inspired by Math Jams. */ | ||
+ | |||
+ | size(300); | ||
+ | pair A, B, C, D, O, P, R, S, T, H, M, N; | ||
+ | A = origin; | ||
+ | B = (125/4,0); | ||
+ | C = B + 125/4 * dir((3,4)); | ||
+ | D = A + 125/4 * dir((3,4)); | ||
+ | O = (25,25/2); | ||
+ | P = (15,5); | ||
+ | R = foot(P,A,D); | ||
+ | S = foot(P,A,B); | ||
+ | T = foot(P,B,C); | ||
+ | H = foot(S,C,D); | ||
+ | M = foot(O,S,H); | ||
+ | N = foot(O,R,T); | ||
+ | |||
+ | fill(O--M--P--cycle,yellow); | ||
+ | fill(O--N--P--cycle,green); | ||
+ | markscalefactor=0.15; | ||
+ | draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red); | ||
+ | draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P); | ||
+ | draw(P--R^^P--S^^P--T^^P--H,red+dashed); | ||
+ | dot("$A$",A,1.5*dir(225),linewidth(4.5)); | ||
+ | dot("$B$",B,1.5*dir(-45),linewidth(4.5)); | ||
+ | dot("$C$",C,1.5*dir(45),linewidth(4.5)); | ||
+ | dot("$D$",D,1.5*dir(90),linewidth(4.5)); | ||
+ | dot("$P$",P,1.5*dir(60),linewidth(4.5)); | ||
+ | dot("$R$",R,1.5*dir(135),linewidth(4.5)); | ||
+ | dot("$S$",S,1.5*dir(-90),linewidth(4.5)); | ||
+ | dot("$T$",T,1.5*dir(-45),linewidth(4.5)); | ||
+ | dot("$O$",O,1.5*dir(45),linewidth(4.5)); | ||
+ | dot("$H$",H,1.5*dir(90),linewidth(4.5)); | ||
+ | dot("$M$",M,1.5*dir(180),linewidth(4.5)); | ||
+ | dot("$N$",N,1.5*dir(15),linewidth(4.5)); | ||
+ | |||
+ | label("$9$",midpoint(P--R),dir(A-D),red); | ||
+ | label("$5$",midpoint(P--S),dir(180),red); | ||
+ | label("$16$",midpoint(P--T),dir(A-D),red); | ||
+ | </asy> | ||
+ | Note that the diameter of <math>\odot O</math> is <math>HS=RT=25,</math> so <math>OP=\frac{25}{2}.</math> It follows that: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>In right <math>\triangle OMP,</math> we have <math>MP=\frac{HS}{2}-PS=\frac{15}{2}</math> by symmetry, from which <math>OM=10</math> by the Pythagorean Theorem.</li><p> | ||
+ | <li>In right <math>\triangle ONP,</math> we have <math>NP=\frac{RT}{2}-RP=\frac{7}{2}</math> by symmetry, from which <math>ON=12</math> by the Pythagorean Theorem.</li><p> | ||
+ | </ol> | ||
+ | Since <math>\overline{MO}\parallel\overline{AB}</math> and <math>\overline{ON}\parallel\overline{DA},</math> we conclude that <math>\angle A = \angle MON.</math> We apply the Sine of a Sum Formula: | ||
+ | <cmath>\begin{align*} | ||
+ | \sin\angle A &= \sin\angle MON \\ | ||
+ | &= \sin(\angle MOP + \angle PON) \\ | ||
+ | &= \sin\angle MOP \cos\angle PON + \cos\angle MOP \sin\angle PON \\ | ||
+ | &= \frac{3}{5}\cdot\frac{24}{25} + \frac{4}{5}\cdot\frac{7}{25} \\ | ||
+ | &= \frac{4}{5}. | ||
+ | \end{align*}</cmath> | ||
+ | Note that <cmath>\sin\angle A = \frac{HS}{DA},</cmath> from which <math>\frac{4}{5} = \frac{25}{DA}.</math> We solve this equation to get <math>DA=\frac{125}{4}.</math> | ||
+ | |||
+ | Finally, the perimeter of <math>ABCD</math> is <math>4DA = \boxed{125}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM (credit given to TheAMCHub) | ||
+ | |||
+ | ==Solution 3== | ||
Label the points of the rhombus to be <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> and the center of the incircle to be <math>O</math> so that <math>9</math>, <math>5</math>, and <math>16</math> are the distances from point <math>P</math> to side <math>ZW</math>, side <math>WX</math>, and <math>XY</math> respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus <math>XYZW</math> is <math>25</math> and circle <math>O</math> has radius <math>\frac{25}{2}</math>. | Label the points of the rhombus to be <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> and the center of the incircle to be <math>O</math> so that <math>9</math>, <math>5</math>, and <math>16</math> are the distances from point <math>P</math> to side <math>ZW</math>, side <math>WX</math>, and <math>XY</math> respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus <math>XYZW</math> is <math>25</math> and circle <math>O</math> has radius <math>\frac{25}{2}</math>. | ||
− | Call the feet of the altitudes from P to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>A</math>, <math>B</math>, and <math>C</math> respectively. Additionally, call the feet of the altitudes from <math>O</math> to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>D</math>, <math>E</math>, and <math>F</math> respectively. | + | Call the feet of the altitudes from <math>P</math> to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>A</math>, <math>B</math>, and <math>C</math> respectively. Additionally, call the feet of the altitudes from <math>O</math> to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>D</math>, <math>E</math>, and <math>F</math> respectively. |
− | Draw a line segment from <math>P</math> to <math>\overline{OD}</math> so that it is perpendicular to <math>\overline{OD}</math>. Notice that this segment length is equal to <math>AD</math> and is <math>\sqrt{(\frac{25}{2})^2-(\frac{7}{2})^2}=12</math> by Pythagorean Theorem. | + | Draw a line segment from <math>P</math> to <math>\overline{OD}</math> so that it is perpendicular to <math>\overline{OD}</math>. Notice that this segment length is equal to <math>AD</math> and is <math>\sqrt{\left(\frac{25}{2}\right)^2-\left(\frac{7}{2}\right)^2}=12</math> by Pythagorean Theorem. |
− | Similarly, perform the same operations with | + | Similarly, perform the same operations with perpendicular from <math>P</math> to <math>\overline{OE}</math> to get <math>BE=10</math>. |
By equal tangents, <math>WD=WE</math>. Now, label the length of segment <math>WA=n</math> and <math>WB=n+2</math>. | By equal tangents, <math>WD=WE</math>. Now, label the length of segment <math>WA=n</math> and <math>WB=n+2</math>. | ||
Line 79: | Line 256: | ||
</cmath> | </cmath> | ||
− | ~Danielzh | + | ~[[Daniel Zhou's Profile|Danielzh]] |
− | ==Solution | + | ==Solution 4== |
Denote by <math>O</math> the center of <math>ABCD</math>. | Denote by <math>O</math> the center of <math>ABCD</math>. | ||
Line 145: | Line 322: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | The center of the incircle is <math>O.</math> Denote the points in which the incircle meets <math>\overline{AB},</math> <math>\overline{BC},</math> <math>\overline{CD},</math> and <math>\overline{DA}</math> as <math>W,</math> <math>X,</math> <math>Y,</math> and <math>Z,</math> respectively. Next, also denote the base of the perpendicular from <math>P</math> to <math>\overline{AB},</math> <math>\overline{AD},</math> <math>\overline{OW},</math> and <math>\overline{OZ}</math> as <math>M,</math> <math>N,</math> <math>S,</math> and <math>T,</math> respectively. | ||
+ | |||
+ | We can easily see that the radius of the circle is <math>\frac{25}{2}.</math> Using this and Pythagorus on right <math>\triangle OSP</math> and <math>\triangle OTP,</math> we find that <math>MW = PS = 10</math> and <math>NZ = PT = 12.</math> | ||
+ | |||
+ | Since <math>AW = AZ</math> by properties of circle tangents, we can deduce by the above information that <math>AM = AN+2.</math> Doing Pythagorus on right <math>\triangle AMP</math> and <math>\triangle ANP</math> we find that <math>a^2 = b^2 + 56</math> (because <math>a^2+25=b^2+81.</math>) From solving the <math>2</math> just derived equations, we find that <math>AM=15</math> and <math>AN=13.</math> | ||
+ | |||
+ | Next, we use Pythagorus on right <math>\triangle AOB</math> (we can see it's right because of properties of rhombuses.) We get <cmath>AB^2 = AO^2 + BO^2.</cmath> We know <math>AB = AW + WB = 25 + WB.</math> By Pythagorus on <math>\triangle AWO</math> and <math>\triangle BWO,</math> we also know <math>AO^2 = 25^2+\left(\frac{25}{2}\right)^2</math> and <math>BO^2=WB^2+\left(\frac{25}{2}\right)^2.</math> Substituting these in, we have <cmath>25^2 + 50WB + WB^2 = 25^2+\left(\frac{25}{2}\right)^2+\left(\frac{25}{2}\right)^2+WB^2.</cmath> Solving for <math>WB,</math> we get <math>WB = \frac{25}{4}.</math> Now we find that each side of the rhombus <math>=AB=25+\frac{25}{4}=\frac{125}{4}.</math> The perimeter of the rhombus would be that times <math>4.</math> Our final answer is <cmath>\frac{125}{4}\cdot4=\boxed{125}.</cmath> | ||
+ | |||
+ | ~s214425 | ||
+ | ==Solution 6 == | ||
+ | [[File:2023 AIME I 8.png|450px|right]] | ||
+ | Notation is shown on diagram, <math>RT \perp AD, FG \perp AB, E = AD \cap \omega, E' = FG \cap AD.</math> | ||
+ | <math>RT = 9 + 16 = 25 = FG</math> as hights of rhombus. | ||
+ | <cmath>RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,</cmath> | ||
+ | <cmath>PE' = 25 - 5 - 5 = 15, RE = \sqrt{RP \cdot RQ} = \sqrt{9 \cdot 16} = 12.</cmath> | ||
+ | <cmath>PE = \sqrt{RP^2 + RE^2} = 15 \implies E = E'.</cmath> | ||
+ | <cmath>\sin \alpha = \frac {RE}{PE} = \frac {GF}{AD} \implies AD = \frac {15 \cdot 25}{12} = \frac {125}{4}.</cmath> | ||
+ | The perimeter of <math>ABCD </math> is <math>\frac{125}{4}\cdot4=\boxed{125}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/AYH6zdJqZLM | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
==See also== | ==See also== |
Latest revision as of 14:54, 3 July 2023
Contents
Problem
Rhombus has
There is a point
on the incircle of the rhombus such that the distances from
to the lines
and
are
and
respectively. Find the perimeter of
Diagram
~MRENTHUSIASM
Solution 1
This solution refers to the Diagram section.
Let be the incenter of
for which
is tangent to
and
at
and
respectively. Moreover, suppose that
and
are the feet of the perpendiculars from
to
and
respectively, such that
intersects
at
and
We obtain the following diagram:
Note that
by the properties of tangents, so
is a rectangle. It follows that the diameter of
is
Let and
We apply the Power of a Point Theorem to
and
We solve this system of equations to get
and
Alternatively, we can find these results by the symmetry on rectangle
and semicircle
We extend beyond
to intersect
and
at
and
respectively, where
So, we have
and
On the other hand, we have
by the Pythagorean Theorem on right
Together, we conclude that
Therefore, points
and
must be collinear.
Let be the foot of the perpendicular from
to
Note that
as shown below:
As
and
by the AA Similarity, we conclude that
The ratio of similitude is
We get
from which
Finally, the perimeter of is
~MRENTHUSIASM (inspired by awesomeming327. and WestSuburb)
Solution 2
This solution refers to the Diagram section.
Define points and
as Solution 1 does. Moreover, let
be the foot of the perpendicular from
to
be the foot of the perpendicular from
to
and
be the foot of the perpendicular from
to
We obtain the following diagram:
Note that the diameter of
is
so
It follows that:
- In right
we have
by symmetry, from which
by the Pythagorean Theorem.
- In right
we have
by symmetry, from which
by the Pythagorean Theorem.
Since and
we conclude that
We apply the Sine of a Sum Formula:
Note that
from which
We solve this equation to get
Finally, the perimeter of is
~MRENTHUSIASM (credit given to TheAMCHub)
Solution 3
Label the points of the rhombus to be ,
,
, and
and the center of the incircle to be
so that
,
, and
are the distances from point
to side
, side
, and
respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus
is
and circle
has radius
.
Call the feet of the altitudes from to side
, side
, and side
to be
,
, and
respectively. Additionally, call the feet of the altitudes from
to side
, side
, and side
to be
,
, and
respectively.
Draw a line segment from to
so that it is perpendicular to
. Notice that this segment length is equal to
and is
by Pythagorean Theorem.
Similarly, perform the same operations with perpendicular from to
to get
.
By equal tangents, . Now, label the length of segment
and
.
Using Pythagorean Theorem again, we get
Which also gives us and
.
Since the diagonals of the rhombus intersect at and are angle bisectors and are also perpendicular to each other, we can get that
Solution 4
Denote by the center of
.
We drop an altitude from
to
that meets
at point
.
We drop altitudes from
to
and
that meet
and
at
and
, respectively.
We denote
.
We denote the side length of
as
.
Because the distances from to
and
are
and
, respectively, and
, the distance between each pair of two parallel sides of
is
.
Thus,
and
.
We have
Thus, .
In , we have
.
Thus,
Taking the imaginary part of this equation and plugging and
into this equation, we get
We have
Because is on the incircle of
,
. Plugging this into
, we get the following equation
By solving this equation, we get and
.
Therefore,
.
Therefore, the perimeter of is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 5
The center of the incircle is Denote the points in which the incircle meets
and
as
and
respectively. Next, also denote the base of the perpendicular from
to
and
as
and
respectively.
We can easily see that the radius of the circle is Using this and Pythagorus on right
and
we find that
and
Since by properties of circle tangents, we can deduce by the above information that
Doing Pythagorus on right
and
we find that
(because
) From solving the
just derived equations, we find that
and
Next, we use Pythagorus on right (we can see it's right because of properties of rhombuses.) We get
We know
By Pythagorus on
and
we also know
and
Substituting these in, we have
Solving for
we get
Now we find that each side of the rhombus
The perimeter of the rhombus would be that times
Our final answer is
~s214425
Solution 6
Notation is shown on diagram,
as hights of rhombus.
The perimeter of
is
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~MathProblemSolvingSkills.com
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
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