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Difference between revisions of "1991 AIME Problems/Problem 12"

(Add four more solutions, by 4everwise, pointios, pkerichang, and ehehheehee)
(→‎Solution: revise other solutions (fix typos etc))
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=== Solution 2 ===
 
=== Solution 2 ===
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From above, we have <math>OB = 24</math> and <math>BD = 48</math>. Returning to <math>BPQO,</math> note that <math>\angle PQO\cong \angle PBO \cong ABD.</math> Hence, <math>\triangle ABD \sim \triangle OQP</math> by <math>AA</math> similarity. From here, it's clear that
+
From above, we have <math>OB = 24</math> and <math>BD = 48</math>. Returning to <math>BPQO,</math> note that <math>\angle PQO\cong \angle PBO \cong ABD.</math> Hence, <math>\triangle ABD \sim \triangle OQP</math> by <math>AA</math> [[similar triangle|similar]]ity. From here, it's clear that
 
<cmath>
 
<cmath>
 
\frac {AD}{BD} = \frac {IP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}.
 
\frac {AD}{BD} = \frac {IP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}.
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=== Solution 3 ===
 
=== Solution 3 ===
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The triangles <math>QOB,OBC</math> are [[isosceles triangle|isosceles]], and similar (because they have <math>\angle QOB = \angle OBC</math>).
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The triangles <math>QOB,OBC</math> are [[isosceles triangle|isosceles]], and [[similar triangles|similar]] (because they have <math>\angle QOB = \angle OBC</math>).
  
 
Hence <math>\frac {BQ}{OB} = \frac {OB}{BC} \Rightarrow OB^2 = BC \cdot BQ</math>
 
Hence <math>\frac {BQ}{OB} = \frac {OB}{BC} \Rightarrow OB^2 = BC \cdot BQ</math>
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=== Solution 4 ===
 
=== Solution 4 ===
−
For convenience, let <math>\angle PQS = \theta</math>. Since the opposite triangles are congruent we have that <math>\angle BQR = 3\theta</math>, and therefore <math>\angle QRC = 3\theta - 90</math>. Let <math>RC = a</math>, then we have <math>\sin{(3\theta - 90)} = \frac {a}{25}</math>, or <math>- \cos{3\theta} = \frac {a}{25}</math>. Expanding with the formula <math>\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}</math>, and since we have <math>\cos{\theta} = \frac {4}{5}</math>, the rest follows similarily to above.
+
For convenience, let <math>\angle PQS = \theta</math>. Since the opposite triangles are congruent we have that <math>\angle BQR = 3\theta</math>, and therefore <math>\angle QRC = 3\theta - 90</math>. Let <math>RC = a</math>, then we have <math>\sin{(3\theta - 90)} = \frac {a}{25}</math>, or <math>- \cos{3\theta} = \frac {a}{25}</math>. Expanding with the formula <math>\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}</math>, and since we have <math>\cos{\theta} = \frac {4}{5}</math>, we can solve for <math>a</math>. The rest then follows similarily from above.
  
 
=== Solution 5 ===
 
=== Solution 5 ===
−
You can just lable the points.  After drawing a brief picture, you can see 4 right triangles with sides of 15,20,25.
+
You can just label the points.  After drawing a brief picture, you can see 4 right triangles with sides of <math>15,\ 20,\ 25</math>.
  
−
So the points of triangle RDS are (0,0) (0,20) and (15,0)
+
Let the points of triangle <math>RDS</math> be <math>(0,0)\ (0,20)\ (15,0)</math>. Since each right triangle can be split into two similar triangles, point <math>(0,0)</math> is <math>12</math> away from the [[hypotenuse]]. By reflecting <math>(0,0)</math> over the hypotenuse, we can get the 3rd point of the second right triangle (aka the intersection of the diagonals of the rhombus) which is <math>(19.2,14.4)</math>.
−
Since each right triangle can be split into two similar triangles, point (0,0) is 12 away from the hypotnuse. By reflecting (0,0) over the hypotnuse, we can get 3rd point of the second right triangle (A.K.A the intersection of the diagonals of the Rhombus) which is (19.2,14.4)
 
  
−
By reflecting (15,0) over diagonal SQ we get P (23.4,28.8). By adding 15 to the x value we get B(38.4,28.8)
+
By reflecting <math>(15,0)</math> over diagonal <math>\overline{SQ}</math> we get <math>P (23.4,28.8)</math>. By adding <math>15</math> to the <math>x</math> value we get <math>B(38.4,28.8)</math>.
  
−
So the perimeter is equal to <math>(38.4 + 28.8)*2</math> which equals <math>\frac {672}{5}</math>.
+
So the perimeter is equal to <math>(38.4 + 28.8)*2 = \frac {672}{5}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:12, 23 October 2007

Problem

Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$, $Q^{}_{}$, $R^{}_{}$, and $S^{}_{}$ are interior points on sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. It is given that $PB^{}_{}=15$, $BQ^{}_{}=20$, $PR^{}_{}=30$, and $QS^{}_{}=40$. Let $m/n^{}_{}$, in lowest terms, denote the perimeter of $ABCD^{}_{}$. Find $m+n^{}_{}$.

Solution

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Solution 1

Let $O$ be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent ($\triangle BPQ \cong \triangle DRS$, $\triangle APS \cong \triangle CRQ$). Quickly we realize that $O$ is also the center of the rectangle.

By the Pythagorean Theorem, we can solve for a side of the rhombus; $PQ = \sqrt{15^2 + 20^2} = 25$. Since the diagonals of a rhombus are perpendicular bisectors, we have that $OP = 15, OQ = 20$. Also, $\angle POQ = 90^{\circ}$, so quadrilateral $BPOQ$ is cyclic. By Ptolemy's Theorem, $25 \cdot OB = 20 \cdot 15 + 15 \cdot 20 = 600$.

By similar logic, we have $APOS$ is a cyclic quadrilateral. Let $AP = x$, $AS = y$. The Pythagorean Theorem gives us $x^2 + y^2 = 625\quad \mathrm{(1)}$. Ptolemy’s Theorem gives us $25 \cdot OA = 20x + 15y$. Since the diagonals of a rectangle are equal, $OA = \frac{1}{2}d = OB$, and $20x + 15y = 600\quad \mathrm{(2)}$. Solving for $y$, we get $y = 40 - \frac 43x$. Substituting into $\mathrm{(1)}$,

\begin{eqnarray*}x^2 + \left(40-\frac 43x\right)^2 &=& 625\\ 5x^2 - 192x + 1755 &=& 0\\ x = \frac{192 \pm \sqrt{192^2 - 4 \cdot 5 \cdot 1755}}{10} &=& 15, \frac{117}{5}\end{eqnarray*}

We reject $15$ because then everything degenerates into squares, but the condition that $PR \neq QS$ gives us a contradiction. Thus $x = \frac{117}{5}$, and backwards solving gives $y = \frac{44}5$. The perimeter of $ABCD$ is $2\left(20 + 15 + \frac{117}{5} + \frac{44}5\right) = \frac{672}{5}$, and $m + n = \boxed{677}$.

Solution 2

From above, we have $OB = 24$ and $BD = 48$. Returning to $BPQO,$ note that $\angle PQO\cong \angle PBO \cong ABD.$ Hence, $\triangle ABD \sim \triangle OQP$ by $AA$ similarity. From here, it's clear that \[\frac {AD}{BD} = \frac {IP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}.\] Similarly, \[\frac {AB}{BD} = \frac {IQ}{PQ}\implies \frac {AB}{48} = \frac {20}{25}\implies AB = \frac {192}{5}.\] Therefore, the perimeter of rectangle $ABCD$ is $2(AB + AD) = 2\left(\frac {192}{5} + \frac {144}{5}\right) = \frac {672}{5}.$

Solution 3

The triangles $QOB,OBC$ are isosceles, and similar (because they have $\angle QOB = \angle OBC$).

Hence $\frac {BQ}{OB} = \frac {OB}{BC} \Rightarrow OB^2 = BC \cdot BQ$

The length of $OB$ could be found easily from the area of $BPQ$:

$BP \cdot PQ = \frac {OB}{2} \cdot PQ \Rightarrow OB = \frac {2BP\cdot PQ}{OB} \Rightarrow OB = 24$

So $OB^2 = BC \cdot BQ \Rightarrow 24^2 = (20 + CQ) \cdot 20 \Rightarrow CQ = \frac {44}{5}$

From the right triangle $CRQ$ we have $RC^2 = 25^2 - (\frac {44}{5})^2\Rightarrow RC = \frac {117}{5}$

$($or we can define a similar formula : $OB^2 = BP \cdot BA$ , and then we find $AP$ in other words the segment $OB$ is tangent to the circles with diameters $AO,CO)$

The perimeter is $2(PB + BQ + QC + CR) = 2(15 + 20 + \frac {44 + 117}{5}) = \frac {672}{5}\Rightarrow m+n=677$.

Solution 4

For convenience, let $\angle PQS = \theta$. Since the opposite triangles are congruent we have that $\angle BQR = 3\theta$, and therefore $\angle QRC = 3\theta - 90$. Let $RC = a$, then we have $\sin{(3\theta - 90)} = \frac {a}{25}$, or $- \cos{3\theta} = \frac {a}{25}$. Expanding with the formula $\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}$, and since we have $\cos{\theta} = \frac {4}{5}$, we can solve for $a$. The rest then follows similarily from above.

Solution 5

You can just label the points. After drawing a brief picture, you can see 4 right triangles with sides of $15,\ 20,\ 25$.

Let the points of triangle $RDS$ be $(0,0)\ (0,20)\ (15,0)$. Since each right triangle can be split into two similar triangles, point $(0,0)$ is $12$ away from the hypotenuse. By reflecting $(0,0)$ over the hypotenuse, we can get the 3rd point of the second right triangle (aka the intersection of the diagonals of the rhombus) which is $(19.2,14.4)$.

By reflecting $(15,0)$ over diagonal $\overline{SQ}$ we get $P (23.4,28.8)$. By adding $15$ to the $x$ value we get $B(38.4,28.8)$.

So the perimeter is equal to $(38.4 + 28.8)*2 = \frac {672}{5}$.

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
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