Difference between revisions of "1991 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | [[Rhombus]] <math>PQRS^{}_{}</math> is [[inscribe]]d in [[rectangle]] <math>ABCD^{}_{}</math> so that [[vertex|vertices]] <math>P^{}_{}</math>, <math>Q^{}_{}</math>, <math>R^{}_{}</math>, and <math>S^{}_{}</math> are interior points on sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, and <math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{}_{}=40</math>. Let <math>m | + | [[Rhombus]] <math>PQRS^{}_{}</math> is [[inscribe]]d in [[rectangle]] <math>ABCD^{}_{}</math> so that [[vertex|vertices]] <math>P^{}_{}</math>, <math>Q^{}_{}</math>, <math>R^{}_{}</math>, and <math>S^{}_{}</math> are interior points on sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, and <math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{}_{}=40</math>. Let <math>\frac{m}{n}</math>, in lowest terms, denote the [[perimeter]] of <math>ABCD^{}_{}</math>. Find <math>m+n^{}_{}</math>. |
__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
− | <center><asy>defaultpen(fontsize( | + | <center><asy>defaultpen(fontsize(12)+linewidth(1.3)); pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q=(38.4,8.8), R=(15,0), S=(0,20); O=intersectionpoint(A--C,B--D); draw(A--B--C--D--cycle);draw(P--R..Q--S); draw(P--Q--R--S--cycle); label(" |
=== Solution 1 === | === Solution 1 === | ||
− | Let <math>O</math> be the center of the rhombus. Via [[parallel]] sides and | + | Let <math>O</math> be the center of the rhombus. Via [[parallel]] sides and alternate interior angles, we see that the opposite [[triangle]]s are [[congruent]] (<math>\triangle BPQ \cong \triangle DRS</math>, <math>\triangle APS \cong \triangle CRQ</math>). Quickly we realize that <math>O</math> is also the center of the rectangle. |
By the [[Pythagorean Theorem]], we can solve for a side of the rhombus; <math>PQ = \sqrt{15^2 + 20^2} = 25</math>. Since the [[diagonal]]s of a rhombus are [[perpendicular bisector]]s, we have that <math>OP = 15, OQ = 20</math>. Also, <math>\angle POQ = 90^{\circ}</math>, so quadrilateral <math>BPOQ</math> is [[cyclic quadrilateral|cyclic]]. By [[Ptolemy's Theorem]], <math>25 \cdot OB = 20 \cdot 15 + 15 \cdot 20 = 600</math>. | By the [[Pythagorean Theorem]], we can solve for a side of the rhombus; <math>PQ = \sqrt{15^2 + 20^2} = 25</math>. Since the [[diagonal]]s of a rhombus are [[perpendicular bisector]]s, we have that <math>OP = 15, OQ = 20</math>. Also, <math>\angle POQ = 90^{\circ}</math>, so quadrilateral <math>BPOQ</math> is [[cyclic quadrilateral|cyclic]]. By [[Ptolemy's Theorem]], <math>25 \cdot OB = 20 \cdot 15 + 15 \cdot 20 = 600</math>. | ||
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=== Solution 6 === | === Solution 6 === | ||
− | We can just use areas. Let <math>AP = b</math> and <math>AS = a</math>. <math>a^2 + b^2 = 625</math>. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, <math>(a+20)(b+15)</math>. This gives <math>3a + 4b = 120</math>. Solving this system of equation gives <math>\frac{44}{5} = a</math>, <math>\frac{117}{5} = b</math>, from which it is straightforward to find the answer, <math>2(a+b+35) \Rightarrow \boxed{677}</math> | + | We can just use areas. Let <math>AP = b</math> and <math>AS = a</math>. <math>a^2 + b^2 = 625</math>. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, <math>(a+20)(b+15)</math>. This gives <math>3a + 4b = 120</math>. Solving this system of equation gives <math>\frac{44}{5} = a</math>, <math>\frac{117}{5} = b</math>, from which it is straightforward to find the answer, <math>2(a+b+35) \Rightarrow \frac{672}{5}</math>. Thus, <math>m+n = \frac{672}{5}\implies\boxed{677}</math> |
=== Solution 7 === | === Solution 7 === |
Latest revision as of 22:36, 27 August 2023
Problem
Rhombus is inscribed in rectangle
so that vertices
,
,
, and
are interior points on sides
,
,
, and
, respectively. It is given that
,
,
, and
. Let
, in lowest terms, denote the perimeter of
. Find
.
Contents
[hide]Solution
![[asy]defaultpen(fontsize(12)+linewidth(1.3)); pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q=(38.4,8.8), R=(15,0), S=(0,20); O=intersectionpoint(A--C,B--D); draw(A--B--C--D--cycle);draw(P--R..Q--S); draw(P--Q--R--S--cycle); label("\(A\)",A,NW);label("\(B\)",B,NE);label("\(C\)",C,SE);label("\(D\)",D,SW); label("\(P\)",P,N);label("\(Q\)",Q,E);label("\(R\)",R,SW);label("\(S\)",S,W); label("\(15\)",B/2+P/2,N);label("\(20\)",B/2+Q/2,E);label("\(O\)",O,SW); [/asy]](http://latex.artofproblemsolving.com/0/a/b/0ab275747a2707450cd73d377348c4cc2543ee20.png)
Solution 1
Let be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent (
,
). Quickly we realize that
is also the center of the rectangle.
By the Pythagorean Theorem, we can solve for a side of the rhombus; . Since the diagonals of a rhombus are perpendicular bisectors, we have that
. Also,
, so quadrilateral
is cyclic. By Ptolemy's Theorem,
.
By similar logic, we have is a cyclic quadrilateral. Let
,
. The Pythagorean Theorem gives us
. Ptolemy’s Theorem gives us
. Since the diagonals of a rectangle are equal,
, and
. Solving for
, we get
. Substituting into
,
We reject because then everything degenerates into squares, but the condition that
gives us a contradiction. Thus
, and backwards solving gives
. The perimeter of
is
, and
.
Solution 2
From above, we have and
. Returning to
note that
Hence,
by
similarity. From here, it's clear that
Similarly,
Therefore, the perimeter of rectangle
is
Solution 3
The triangles are isosceles, and similar (because they have
).
Hence .
The length of could be found easily from the area of
:
From the right triangle we have
. We could have also defined a similar formula:
, and then we found
, the segment
is tangent to the circles with diameters
.
The perimeter is .
Solution 4
For convenience, let . Since the opposite triangles are congruent we have that
, and therefore
. Let
, then we have
, or
. Expanding with the formula
, and since we have
, we can solve for
. The rest then follows similarily from above.
Solution 5
We can just find coordinates of the points. After drawing a picture, we can see 4 congruent right triangles with sides of , namely triangles
and
.
Let the points of triangle be
. Let point
be on
, such that
and
. Triangle
can be split into two similar 3-4-5 right triangles,
and
. By the Pythagorean Theorem, point
is
away from point
. Repeating the process, if we break down triangle
into two more similar triangles, we find that point
is at
.
By reflecting point over point
, we get point
. By reflecting point
over point
, we get point
. Thus, the perimeter is equal to
, making the final answer
.
Solution 6
We can just use areas. Let and
.
. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle,
. This gives
. Solving this system of equation gives
,
, from which it is straightforward to find the answer,
. Thus,
Solution 7
We will bash with trigonometry.
Firstly, by Pythagoras Theorem, . We observe that
. Thus, if we drop an altitude from
to
to point
, it will have length
. In particular,
since we form a 7-24-25 triangle.
Now, . Thus, since
, we get that
. Now, by the Pythagorean Theorem,
.
Using the same idea, . Thus, since
.
Now, we can finish. We know . We also know
. Thus, our perimeter is
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.