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Difference between revisions of "2021 AIME I Problems/Problem 9"

(→‎Solution 7 (Similar Triangles))
(Made the titles consistent.)
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2 (Similar Triangles and Trigonometry)==
+
==Solution 1 (Similar Triangles)==
−
Let <math>AD=BC=a</math>. Draw diagonal <math>AC</math> and let <math>G</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>, <math>F</math> be the foot of the perpendicular from <math>A</math> to line <math>BC</math>, and <math>H</math> be the foot of the perpendicular from <math>A</math> to <math>DC</math>.
 
−
 
 
−
Note that <math>\triangle CBG\sim\triangle CAF</math>, and we get that <math>\frac{10}{15}=\frac{a}{AC}</math>. Therefore, <math>AC=\frac32 a</math>. It then follows that <math>\triangle ABF\sim\triangle ADH</math>. Using similar triangles, we can then find that <math>AB=\frac{5}{6}a</math>. Using the Law of Cosines on <math>\triangle ABC</math>, We can find that the <math>\cos\angle ABC=-\frac{1}{3}</math>. Since <math>\angle ABF=\angle ADH</math>, and each is supplementary to <math>\angle ABC</math>, we know that the <math>\cos\angle ADH=\frac{1}{3}</math>. It then follows that <math>a=\frac{27\sqrt{2}}{2}</math>. Then it can be found that the area <math>K</math> is <math>\frac{567\sqrt{2}}{2}</math>. Multiplying this by <math>\sqrt{2}</math>, the answer is <math>\boxed{567}</math>.
 
−
 
 
−
~happykeeper
 
−
 
 
−
==Solution 3 (Similar Triangles and Cyclic Quadrilaterals)==
 
−
Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>P</math>, to <math>CD</math> be <math>Q</math>, and to <math>BD</math> be <math>R</math>.
 
−
 
 
−
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, <math>A</math> is on the circumcircle of <math>\triangle BCD</math> and we have that <math>PRQ</math> is the Simson Line from <math>A</math>. As <math>\angle QAB = 90^\circ</math>, we have that <math>\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ</math>, with the last equality coming from cyclic quadrilateral <math>APBR</math>. Thus, <math>\triangle QAR \sim \triangle QPA</math> and we have that <math>\frac{AQ}{AR} = \frac{PQ}{PA}</math> or that <math>\frac{18}{10} = \frac{QP}{15}</math>, which we can see gives us that <math>QP = 27</math>. Further ratios using the same similar triangles gives that <math>QR = \frac{25}{3}</math> and <math>RP = \frac{56}{3}</math>.
 
−
 
 
−
We also see that quadrilaterals <math>APBR</math> and <math>ARDQ</math> are both cyclic, with diameters of the circumcircles being <math>AB</math> and <math>AQ</math> respectively. The intersection of the circumcircles are the points <math>A</math> and <math>R</math>, and we know <math>DRB</math> and <math>QRP</math> are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center <math>A</math> taking <math>\triangle APQ</math> to <math>\triangle APD</math>. Because we know a lot about <math>\triangle APQ</math> but very little about <math>\triangle APD</math> and we would like to know more, we wish to find the ratio of similitude between the two triangles.
 
−
 
 
−
To do this, we use the one number we have for <math>\triangle APD</math>: we know that the altitude from <math>A</math> to <math>BD</math> has length <math>10</math>. As the two triangles are similar, if we can find the height from <math>A</math> to <math>PQ</math>, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that <math>QP = 27</math>. Using this, we can drop the altitude from <math>A</math> to <math>QP</math> and let it intersect <math>QP</math> at <math>H</math>. Then, let <math>QH = x</math> and thus <math>HP=27-x</math>. We then have by the Pythagorean Theorem on <math>\triangle AQH</math> and <math>\triangle APH</math>:
 
−
<cmath>\begin{align*}
 
−
15^2 - x^2 &= 18^2 - (27-x)^2 \\
 
−
225 - x^2 &= 324 - (x^2-54x+729) \\
 
−
54x &= 630 \\
 
−
x &= \frac{35}{3}.
 
−
\end{align*}</cmath>
 
−
Then, <math>RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}</math>. This gives us then from right triangle <math>\triangle ARH</math> that <math>AH = \frac{20\sqrt{2}}{3}</math> and thus the ratio of <math>\triangle APQ</math> to <math>\triangle ABD</math> is <math>\frac{3\sqrt{2}}{4}</math>. From this, we see then that <cmath>AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}</cmath> and <cmath>AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.</cmath> The Pythagorean Theorem on <math>\triangle AQD</math> then gives that <cmath>QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.</cmath>
 
−
Then, we have the height of trapezoid <math>ABCD</math> is <math>AQ = 18</math>, the top base is <math>AB = \frac{45\sqrt{2}}{4}</math>, and the bottom base is <math>CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}</math>. From the equation of a trapezoid, <math>K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}</math>, so the answer is <math>K\sqrt{2} = \boxed{567}</math>.
 
−
 
 
−
~lvmath
 
−
 
 
−
==Solution 4 (Similar Triangles)==
 
−
 
 
−
First, draw the diagram. Then, notice that since <math>ABCD</math> is isosceles, <math>\Delta ABD \cong \Delta BAC</math>, and the length of the altitude from <math>B</math> to <math>AC</math> is also <math>10</math>. Let the foot of this altitude be <math>F</math>, and let the foot of the altitude from <math>A</math> to <math>BC</math> be denoted as <math>E</math>. Then, <math>\Delta BCF \sim \Delta ACE</math>. So, <math>\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}</math>. Now, notice that <math>[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}</math>, where <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Letting <math>AB = x</math>, this equality becomes <math>AC = \frac{9x}{5}</math>. Also, from <math>\frac{BC}{AC} = \frac{2}{3}</math>, we have <math>BC = \frac{6x}{5}</math>. Now, by the Pythagorean theorem on triangles <math>ABF</math> and <math>CBF</math>, we have <math>AF = \sqrt{x^{2}-100}</math> and <math>CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}</math>. Notice that <math>AC = AF + CF</math>, so <math>\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}</math>. Squaring both sides of the equation once, moving <math>x^{2}-100</math> and <math> \left( \frac{6x}{5} \right) ^{2}-100</math> to the right, dividing both sides by <math>2</math>, and squaring the equation once more, we are left with <math>\frac{32x^{4}}{25} = 324x^{2}</math>. Dividing both sides by <math>x^{2}</math> (since we know <math>x</math> is positive), we are left with <math>\frac{32x^{2}}{25} = 324</math>. Solving for <math>x</math> gives us <math>x = \frac{45}{2\sqrt{2}}</math>.
 
−
 
 
−
Now, let the foot of the perpendicular from <math>A</math> to <math>CD</math> be <math>G</math>. Then let <math>DG = y</math>. Let the foot of the perpendicular from <math>B</math> to <math>CD</math> be <math>H</math>. Then, <math>CH</math> is also equal to <math>y</math>. Notice that <math>ABHG</math> is a rectangle, so <math>GH = x</math>. Now, we have <math>CG = GH + CH = x + y</math>. By the Pythagorean theorem applied to <math>\Delta AGC</math>, we have <math>(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}</math>. We know that <math>\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}</math>, so we can plug this into this equation. Solving for <math>x+y</math>, we get <math>x+y=\frac{63}{2\sqrt{2}}</math>.
 
−
 
 
−
Finally, to find <math>[ABCD]</math>, we use the formula for the area of a trapezoid: <math>K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}</math>. The problem asks us for <math>K \cdot \sqrt{2}</math>, which comes out to be <math>\boxed{567}</math>.
 
−
 
 
−
~advanture
 
−
 
 
−
==Solution 5 (Similar Triangles and Cyclic Quadrilaterals)==
 
−
 
 
−
Let <math>E,F,</math> and <math>G</math> be the feet of the altitudes from <math>A</math> to <math>BC,CD,</math> and <math>DB</math>, respectively.
 
−
 
 
−
Claim: We have <math>2</math> pairs of similar right triangles: <math>\triangle AEB \sim \triangle AFD</math> and <math>\triangle AGD \sim \triangle AEC</math>.
 
−
 
 
−
Proof: Note that <math>ABCD</math> is cyclic. We need one more angle, and we get this from this cyclic quadrilateral:
 
−
<cmath>\begin{align*}
 
−
\angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\
 
−
\angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square
 
−
\end{align*}</cmath>
 
−
Let <math>AD=a</math>. We obtain from the similarities <math>AB = \frac{5a}{6}</math> and <math>AC=BD=\frac{3a}{2}</math>.
 
−
 
 
−
By Ptolemy, <math>\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD</math>, so <math>\frac{5a^2}{4} = \frac{5a}{6} \cdot CD</math>.
 
−
 
 
−
We obtain <math>CD=\frac{3a}{2}</math>, so <math>DF=\frac{CD-AB}{2}=\frac{a}{3}</math>.
 
−
 
 
−
Applying the Pythagorean theorem on <math>\triangle ADF</math>, we get <math>324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}</math>.
 
−
 
 
−
Thus, <math>a=\frac{27}{\sqrt{2}}</math>, and <math>[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}</math>, yielding <math>\sqrt2\cdot[ABCD]=\boxed{567}</math>.
 
−
 
 
−
==Solution 6 (Similar Triangles)==
 
 
Let <math>\overline{AE}, \overline{AF},</math> and <math>\overline{AG}</math> be the perpendiculars from <math>A</math> to <math>\overleftrightarrow{BC}, \overleftrightarrow{CD},</math> and <math>\overleftrightarrow{BD},</math> respectively. Next, let <math>H</math> be the intersection of <math>\overline{AF}</math> and <math>\overline{BD}.</math>
 
Let <math>\overline{AE}, \overline{AF},</math> and <math>\overline{AG}</math> be the perpendiculars from <math>A</math> to <math>\overleftrightarrow{BC}, \overleftrightarrow{CD},</math> and <math>\overleftrightarrow{BD},</math> respectively. Next, let <math>H</math> be the intersection of <math>\overline{AF}</math> and <math>\overline{BD}.</math>
  
Line 141: Line 84:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 7 (Similar Triangles and Trigonometry)==
+
==Solution 2 (Similar Triangles)==
 +
 
 +
First, draw the diagram. Then, notice that since <math>ABCD</math> is isosceles, <math>\Delta ABD \cong \Delta BAC</math>, and the length of the altitude from <math>B</math> to <math>AC</math> is also <math>10</math>. Let the foot of this altitude be <math>F</math>, and let the foot of the altitude from <math>A</math> to <math>BC</math> be denoted as <math>E</math>. Then, <math>\Delta BCF \sim \Delta ACE</math>. So, <math>\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}</math>. Now, notice that <math>[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}</math>, where <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Letting <math>AB = x</math>, this equality becomes <math>AC = \frac{9x}{5}</math>. Also, from <math>\frac{BC}{AC} = \frac{2}{3}</math>, we have <math>BC = \frac{6x}{5}</math>. Now, by the Pythagorean theorem on triangles <math>ABF</math> and <math>CBF</math>, we have <math>AF = \sqrt{x^{2}-100}</math> and <math>CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}</math>. Notice that <math>AC = AF + CF</math>, so <math>\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}</math>. Squaring both sides of the equation once, moving <math>x^{2}-100</math> and <math> \left( \frac{6x}{5} \right) ^{2}-100</math> to the right, dividing both sides by <math>2</math>, and squaring the equation once more, we are left with <math>\frac{32x^{4}}{25} = 324x^{2}</math>. Dividing both sides by <math>x^{2}</math> (since we know <math>x</math> is positive), we are left with <math>\frac{32x^{2}}{25} = 324</math>. Solving for <math>x</math> gives us <math>x = \frac{45}{2\sqrt{2}}</math>.
 +
 
 +
Now, let the foot of the perpendicular from <math>A</math> to <math>CD</math> be <math>G</math>. Then let <math>DG = y</math>. Let the foot of the perpendicular from <math>B</math> to <math>CD</math> be <math>H</math>. Then, <math>CH</math> is also equal to <math>y</math>. Notice that <math>ABHG</math> is a rectangle, so <math>GH = x</math>. Now, we have <math>CG = GH + CH = x + y</math>. By the Pythagorean theorem applied to <math>\Delta AGC</math>, we have <math>(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}</math>. We know that <math>\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}</math>, so we can plug this into this equation. Solving for <math>x+y</math>, we get <math>x+y=\frac{63}{2\sqrt{2}}</math>.
 +
 
 +
Finally, to find <math>[ABCD]</math>, we use the formula for the area of a trapezoid: <math>K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}</math>. The problem asks us for <math>K \cdot \sqrt{2}</math>, which comes out to be <math>\boxed{567}</math>.
 +
 
 +
~advanture
 +
 
 +
==Solution 3 (Similar Triangles and Cyclic Quadrilaterals)==
 +
Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>P</math>, to <math>CD</math> be <math>Q</math>, and to <math>BD</math> be <math>R</math>.
 +
 
 +
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, <math>A</math> is on the circumcircle of <math>\triangle BCD</math> and we have that <math>PRQ</math> is the Simson Line from <math>A</math>. As <math>\angle QAB = 90^\circ</math>, we have that <math>\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ</math>, with the last equality coming from cyclic quadrilateral <math>APBR</math>. Thus, <math>\triangle QAR \sim \triangle QPA</math> and we have that <math>\frac{AQ}{AR} = \frac{PQ}{PA}</math> or that <math>\frac{18}{10} = \frac{QP}{15}</math>, which we can see gives us that <math>QP = 27</math>. Further ratios using the same similar triangles gives that <math>QR = \frac{25}{3}</math> and <math>RP = \frac{56}{3}</math>.
 +
 
 +
We also see that quadrilaterals <math>APBR</math> and <math>ARDQ</math> are both cyclic, with diameters of the circumcircles being <math>AB</math> and <math>AQ</math> respectively. The intersection of the circumcircles are the points <math>A</math> and <math>R</math>, and we know <math>DRB</math> and <math>QRP</math> are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center <math>A</math> taking <math>\triangle APQ</math> to <math>\triangle APD</math>. Because we know a lot about <math>\triangle APQ</math> but very little about <math>\triangle APD</math> and we would like to know more, we wish to find the ratio of similitude between the two triangles.
 +
 
 +
To do this, we use the one number we have for <math>\triangle APD</math>: we know that the altitude from <math>A</math> to <math>BD</math> has length <math>10</math>. As the two triangles are similar, if we can find the height from <math>A</math> to <math>PQ</math>, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that <math>QP = 27</math>. Using this, we can drop the altitude from <math>A</math> to <math>QP</math> and let it intersect <math>QP</math> at <math>H</math>. Then, let <math>QH = x</math> and thus <math>HP=27-x</math>. We then have by the Pythagorean Theorem on <math>\triangle AQH</math> and <math>\triangle APH</math>:
 +
<cmath>\begin{align*}
 +
15^2 - x^2 &= 18^2 - (27-x)^2 \\
 +
225 - x^2 &= 324 - (x^2-54x+729) \\
 +
54x &= 630 \\
 +
x &= \frac{35}{3}.
 +
\end{align*}</cmath>
 +
Then, <math>RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}</math>. This gives us then from right triangle <math>\triangle ARH</math> that <math>AH = \frac{20\sqrt{2}}{3}</math> and thus the ratio of <math>\triangle APQ</math> to <math>\triangle ABD</math> is <math>\frac{3\sqrt{2}}{4}</math>. From this, we see then that <cmath>AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}</cmath> and <cmath>AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.</cmath> The Pythagorean Theorem on <math>\triangle AQD</math> then gives that <cmath>QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.</cmath>
 +
Then, we have the height of trapezoid <math>ABCD</math> is <math>AQ = 18</math>, the top base is <math>AB = \frac{45\sqrt{2}}{4}</math>, and the bottom base is <math>CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}</math>. From the equation of a trapezoid, <math>K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}</math>, so the answer is <math>K\sqrt{2} = \boxed{567}</math>.
 +
 
 +
~lvmath
 +
 
 +
==Solution 4 (Similar Triangles and Cyclic Quadrilaterals)==
 +
 
 +
Let <math>E,F,</math> and <math>G</math> be the feet of the altitudes from <math>A</math> to <math>BC,CD,</math> and <math>DB</math>, respectively.
 +
 
 +
Claim: We have <math>2</math> pairs of similar right triangles: <math>\triangle AEB \sim \triangle AFD</math> and <math>\triangle AGD \sim \triangle AEC</math>.
 +
 
 +
Proof: Note that <math>ABCD</math> is cyclic. We need one more angle, and we get this from this cyclic quadrilateral:
 +
<cmath>\begin{align*}
 +
\angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\
 +
\angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square
 +
\end{align*}</cmath>
 +
Let <math>AD=a</math>. We obtain from the similarities <math>AB = \frac{5a}{6}</math> and <math>AC=BD=\frac{3a}{2}</math>.
 +
 
 +
By Ptolemy, <math>\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD</math>, so <math>\frac{5a^2}{4} = \frac{5a}{6} \cdot CD</math>.
 +
 
 +
We obtain <math>CD=\frac{3a}{2}</math>, so <math>DF=\frac{CD-AB}{2}=\frac{a}{3}</math>.
 +
 
 +
Applying the Pythagorean theorem on <math>\triangle ADF</math>, we get <math>324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}</math>.
 +
 
 +
Thus, <math>a=\frac{27}{\sqrt{2}}</math>, and <math>[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}</math>, yielding <math>\sqrt2\cdot[ABCD]=\boxed{567}</math>.
 +
 
 +
==Solution 5 (Similar Triangles and Trigonometry)==
 +
Let <math>AD=BC=a</math>. Draw diagonal <math>AC</math> and let <math>G</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>, <math>F</math> be the foot of the perpendicular from <math>A</math> to line <math>BC</math>, and <math>H</math> be the foot of the perpendicular from <math>A</math> to <math>DC</math>.
 +
 
 +
Note that <math>\triangle CBG\sim\triangle CAF</math>, and we get that <math>\frac{10}{15}=\frac{a}{AC}</math>. Therefore, <math>AC=\frac32 a</math>. It then follows that <math>\triangle ABF\sim\triangle ADH</math>. Using similar triangles, we can then find that <math>AB=\frac{5}{6}a</math>. Using the Law of Cosines on <math>\triangle ABC</math>, We can find that the <math>\cos\angle ABC=-\frac{1}{3}</math>. Since <math>\angle ABF=\angle ADH</math>, and each is supplementary to <math>\angle ABC</math>, we know that the <math>\cos\angle ADH=\frac{1}{3}</math>. It then follows that <math>a=\frac{27\sqrt{2}}{2}</math>. Then it can be found that the area <math>K</math> is <math>\frac{567\sqrt{2}}{2}</math>. Multiplying this by <math>\sqrt{2}</math>, the answer is <math>\boxed{567}</math>.
 +
 
 +
~happykeeper
 +
 
 +
==Solution 6 (Similar Triangles and Trigonometry)==
 
Draw the distances in terms of <math>B</math>, as shown in the diagram. By similar triangles, <math>\triangle{AEC}\sim\triangle{BIC}</math>. As a result, let <math>AB=u</math>, then <math>BC=AD=\frac{6}{5}u</math> and <math>2AC=3BC</math>. The triangle <math>ABC</math> is <math>6-5-9</math> which <math>\cos(\angle{ABC})=-\frac{1}{3}</math>. By angle subtraction, <math>\cos(180-\theta)=-\cos\theta</math>. Therefore, <math>AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}</math> and <math>AD=BC=\frac{27}{\sqrt{2}}</math>. By trapezoid area formula, the area of <math>ABCD</math> is equal to <math>(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}</math> which <math>\sqrt{2}\cdot k=\boxed{567}</math>.
 
Draw the distances in terms of <math>B</math>, as shown in the diagram. By similar triangles, <math>\triangle{AEC}\sim\triangle{BIC}</math>. As a result, let <math>AB=u</math>, then <math>BC=AD=\frac{6}{5}u</math> and <math>2AC=3BC</math>. The triangle <math>ABC</math> is <math>6-5-9</math> which <math>\cos(\angle{ABC})=-\frac{1}{3}</math>. By angle subtraction, <math>\cos(180-\theta)=-\cos\theta</math>. Therefore, <math>AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}</math> and <math>AD=BC=\frac{27}{\sqrt{2}}</math>. By trapezoid area formula, the area of <math>ABCD</math> is equal to <math>(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}</math> which <math>\sqrt{2}\cdot k=\boxed{567}</math>.
  

Revision as of 12:10, 13 December 2021

Problem

Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

Diagram

[asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); [/asy] ~MRENTHUSIASM

Solution 1 (Similar Triangles)

Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$

We set $AB=x$ and $AH=y,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*S,linewidth(4)); dot("$G$",G,SE,linewidth(4)); dot("$H$",H,SE,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); label("$x$",midpoint(A--B),N); label("$y$",midpoint(A--H),W); [/asy] From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\sqrt{x^2-10^2}$ and $HG=\sqrt{y^2-10^2}$ by the Pythagorean Theorem.

Since $\angle ABG$ and $\angle HAG$ are both complementary to $\angle AHB,$ we have $\angle ABG = \angle HAG,$ from which $\triangle ABG \sim \triangle HAG$ by AA. It follows that $\frac{BG}{AG}=\frac{AG}{HG},$ so $BG\cdot HG=AG^2,$ or \[\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)\] Since $\angle AHB = \angle FHD$ by vertical angles, we have $\triangle AHB \sim \triangle FHD$ by AA, with the ratio of similitude $\frac{AH}{FH}=\frac{BA}{DF}.$ It follows that $DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.$

Since $\angle EBA = \angle ECD = \angle FDA$ by angle chasing, we have $\triangle EBA \sim \triangle FDA$ by AA, with the ratio of similitude $\frac{EA}{FA}=\frac{BA}{DA}.$ It follows that $DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.$

By the Pythagorean Theorem on right $\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \[\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)\] Solving this system of equations ($(1)$ and $(2)$), we get $x=\frac{45\sqrt2}{4}$ and $y=\frac{90}{7},$ so $AB=x=\frac{45\sqrt2}{4}$ and $CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.$ Finally, the area of $ABCD$ is \[K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},\] from which $\sqrt2 \cdot K=\boxed{567}.$

~MRENTHUSIASM

Solution 2 (Similar Triangles)

First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\Delta ABD \cong \Delta BAC$, and the length of the altitude from $B$ to $AC$ is also $10$. Let the foot of this altitude be $F$, and let the foot of the altitude from $A$ to $BC$ be denoted as $E$. Then, $\Delta BCF \sim \Delta ACE$. So, $\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}$. Now, notice that $[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}$, where $[ABC]$ denotes the area of triangle $ABC$. Letting $AB = x$, this equality becomes $AC = \frac{9x}{5}$. Also, from $\frac{BC}{AC} = \frac{2}{3}$, we have $BC = \frac{6x}{5}$. Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$, we have $AF = \sqrt{x^{2}-100}$ and $CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$. Notice that $AC = AF + CF$, so $\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$. Squaring both sides of the equation once, moving $x^{2}-100$ and $\left( \frac{6x}{5} \right) ^{2}-100$ to the right, dividing both sides by $2$, and squaring the equation once more, we are left with $\frac{32x^{4}}{25} = 324x^{2}$. Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\frac{32x^{2}}{25} = 324$. Solving for $x$ gives us $x = \frac{45}{2\sqrt{2}}$.

Now, let the foot of the perpendicular from $A$ to $CD$ be $G$. Then let $DG = y$. Let the foot of the perpendicular from $B$ to $CD$ be $H$. Then, $CH$ is also equal to $y$. Notice that $ABHG$ is a rectangle, so $GH = x$. Now, we have $CG = GH + CH = x + y$. By the Pythagorean theorem applied to $\Delta AGC$, we have $(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}$. We know that $\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}$, so we can plug this into this equation. Solving for $x+y$, we get $x+y=\frac{63}{2\sqrt{2}}$.

Finally, to find $[ABCD]$, we use the formula for the area of a trapezoid: $K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}$. The problem asks us for $K \cdot \sqrt{2}$, which comes out to be $\boxed{567}$.

~advanture

Solution 3 (Similar Triangles and Cyclic Quadrilaterals)

Let the foot of the altitude from $A$ to $BC$ be $P$, to $CD$ be $Q$, and to $BD$ be $R$.

Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\triangle BCD$ and we have that $PRQ$ is the Simson Line from $A$. As $\angle QAB = 90^\circ$, we have that $\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ$, with the last equality coming from cyclic quadrilateral $APBR$. Thus, $\triangle QAR \sim \triangle QPA$ and we have that $\frac{AQ}{AR} = \frac{PQ}{PA}$ or that $\frac{18}{10} = \frac{QP}{15}$, which we can see gives us that $QP = 27$. Further ratios using the same similar triangles gives that $QR = \frac{25}{3}$ and $RP = \frac{56}{3}$.

We also see that quadrilaterals $APBR$ and $ARDQ$ are both cyclic, with diameters of the circumcircles being $AB$ and $AQ$ respectively. The intersection of the circumcircles are the points $A$ and $R$, and we know $DRB$ and $QRP$ are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center $A$ taking $\triangle APQ$ to $\triangle APD$. Because we know a lot about $\triangle APQ$ but very little about $\triangle APD$ and we would like to know more, we wish to find the ratio of similitude between the two triangles.

To do this, we use the one number we have for $\triangle APD$: we know that the altitude from $A$ to $BD$ has length $10$. As the two triangles are similar, if we can find the height from $A$ to $PQ$, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that $QP = 27$. Using this, we can drop the altitude from $A$ to $QP$ and let it intersect $QP$ at $H$. Then, let $QH = x$ and thus $HP=27-x$. We then have by the Pythagorean Theorem on $\triangle AQH$ and $\triangle APH$: \begin{align*} 15^2 - x^2 &= 18^2 - (27-x)^2 \\ 225 - x^2 &= 324 - (x^2-54x+729) \\ 54x &= 630 \\ x &= \frac{35}{3}. \end{align*} Then, $RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}$. This gives us then from right triangle $\triangle ARH$ that $AH = \frac{20\sqrt{2}}{3}$ and thus the ratio of $\triangle APQ$ to $\triangle ABD$ is $\frac{3\sqrt{2}}{4}$. From this, we see then that \[AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}\] and \[AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.\] The Pythagorean Theorem on $\triangle AQD$ then gives that \[QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.\] Then, we have the height of trapezoid $ABCD$ is $AQ = 18$, the top base is $AB = \frac{45\sqrt{2}}{4}$, and the bottom base is $CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}$. From the equation of a trapezoid, $K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}$, so the answer is $K\sqrt{2} = \boxed{567}$.

~lvmath

Solution 4 (Similar Triangles and Cyclic Quadrilaterals)

Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$, respectively.

Claim: We have $2$ pairs of similar right triangles: $\triangle AEB \sim \triangle AFD$ and $\triangle AGD \sim \triangle AEC$.

Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: \begin{align*} \angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\ \angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square \end{align*} Let $AD=a$. We obtain from the similarities $AB = \frac{5a}{6}$ and $AC=BD=\frac{3a}{2}$.

By Ptolemy, $\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD$, so $\frac{5a^2}{4} = \frac{5a}{6} \cdot CD$.

We obtain $CD=\frac{3a}{2}$, so $DF=\frac{CD-AB}{2}=\frac{a}{3}$.

Applying the Pythagorean theorem on $\triangle ADF$, we get $324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}$.

Thus, $a=\frac{27}{\sqrt{2}}$, and $[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}$, yielding $\sqrt2\cdot[ABCD]=\boxed{567}$.

Solution 5 (Similar Triangles and Trigonometry)

Let $AD=BC=a$. Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$, $F$ be the foot of the perpendicular from $A$ to line $BC$, and $H$ be the foot of the perpendicular from $A$ to $DC$.

Note that $\triangle CBG\sim\triangle CAF$, and we get that $\frac{10}{15}=\frac{a}{AC}$. Therefore, $AC=\frac32 a$. It then follows that $\triangle ABF\sim\triangle ADH$. Using similar triangles, we can then find that $AB=\frac{5}{6}a$. Using the Law of Cosines on $\triangle ABC$, We can find that the $\cos\angle ABC=-\frac{1}{3}$. Since $\angle ABF=\angle ADH$, and each is supplementary to $\angle ABC$, we know that the $\cos\angle ADH=\frac{1}{3}$. It then follows that $a=\frac{27\sqrt{2}}{2}$. Then it can be found that the area $K$ is $\frac{567\sqrt{2}}{2}$. Multiplying this by $\sqrt{2}$, the answer is $\boxed{567}$.

~happykeeper

Solution 6 (Similar Triangles and Trigonometry)

Draw the distances in terms of $B$, as shown in the diagram. By similar triangles, $\triangle{AEC}\sim\triangle{BIC}$. As a result, let $AB=u$, then $BC=AD=\frac{6}{5}u$ and $2AC=3BC$. The triangle $ABC$ is $6-5-9$ which $\cos(\angle{ABC})=-\frac{1}{3}$. By angle subtraction, $\cos(180-\theta)=-\cos\theta$. Therefore, $AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}$ and $AD=BC=\frac{27}{\sqrt{2}}$. By trapezoid area formula, the area of $ABCD$ is equal to $(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}$ which $\sqrt{2}\cdot k=\boxed{567}$.

~math2718281828459

Video Solution

https://www.youtube.com/watch?v=6rLnl8z7lnM

See Also

2021 AIME I (Problems • Answer Key • Resources)
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