Difference between revisions of "2021 AIME I Problems/Problem 9"
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==Problem== | ==Problem== | ||
Let <math>ABCD</math> be an isosceles trapezoid with <math>AD=BC</math> and <math>AB<CD.</math> Suppose that the distances from <math>A</math> to the lines <math>BC,CD,</math> and <math>BD</math> are <math>15,18,</math> and <math>10,</math> respectively. Let <math>K</math> be the area of <math>ABCD.</math> Find <math>\sqrt2 \cdot K.</math> | Let <math>ABCD</math> be an isosceles trapezoid with <math>AD=BC</math> and <math>AB<CD.</math> Suppose that the distances from <math>A</math> to the lines <math>BC,CD,</math> and <math>BD</math> are <math>15,18,</math> and <math>10,</math> respectively. Let <math>K</math> be the area of <math>ABCD.</math> Find <math>\sqrt2 \cdot K.</math> | ||
+ | |||
+ | ==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== | ==Solution 1== | ||
Construct your isosceles trapezoid. Let, for simplicity, <math>AB = a</math>, <math>AD = BC = b</math>, and <math>CD = c</math>. Extend the sides <math>BC</math> and <math>AD</math> mark the intersection as <math>P</math>. Following what the question states, drop a perpendicular from <math>A</math> to <math>BC</math> labeling the foot as <math>G</math>. Drop another perpendicular from <math>A</math> to <math>CD</math>, calling the foot <math>E</math>. Lastly, drop a perpendicular from <math>A</math> to <math>BD</math>, labeling it <math>F</math>. In addition, drop a perpendicular from <math>B</math> to <math>AC</math> calling its foot <math>F'</math>. | Construct your isosceles trapezoid. Let, for simplicity, <math>AB = a</math>, <math>AD = BC = b</math>, and <math>CD = c</math>. Extend the sides <math>BC</math> and <math>AD</math> mark the intersection as <math>P</math>. Following what the question states, drop a perpendicular from <math>A</math> to <math>BC</math> labeling the foot as <math>G</math>. Drop another perpendicular from <math>A</math> to <math>CD</math>, calling the foot <math>E</math>. Lastly, drop a perpendicular from <math>A</math> to <math>BD</math>, labeling it <math>F</math>. In addition, drop a perpendicular from <math>B</math> to <math>AC</math> calling its foot <math>F'</math>. | ||
− | |||
− | |||
Start out by constructing a triangle <math>ADH</math> congruent to <math>\triangle ABC</math> with its side of length <math>a</math> on line <math>DE</math>. This works because all isosceles triangles are cyclic and as a result, <math>\angle ADC + \angle ABC = 180^\circ</math>. | Start out by constructing a triangle <math>ADH</math> congruent to <math>\triangle ABC</math> with its side of length <math>a</math> on line <math>DE</math>. This works because all isosceles triangles are cyclic and as a result, <math>\angle ADC + \angle ABC = 180^\circ</math>. | ||
Line 20: | Line 51: | ||
~Math_Genius_164 | ~Math_Genius_164 | ||
− | ==Solution 2(LOC and Trig)== | + | |
− | + | ==Solution 2 (LOC and Trig)== | |
+ | 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 | -happykeeper | ||
− | ==See | + | ==Solution 3 (Similarity)== |
+ | 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 (Cool Solution by advanture)== | ||
+ | |||
+ | 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 (Compact Similarity Solution)== | ||
+ | |||
+ | 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 quad: <cmath>\angle ABE = 180^\circ - \angle ABC =\angle ADC = \angle ADG</cmath> | ||
+ | <cmath>\angle ADG = \angle ADB =\angle ACB = \angle ACE \square </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>\boxed{567}</math>. | ||
+ | |||
+ | ==Solution 6 (Two Variables, Two Equations)== | ||
+ | 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> | ||
+ | |||
+ | We set <math>AB=x</math> and <math>AH=y,</math> 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 <math>HF=18-y</math> by segment subtraction, and <math>BG=\sqrt{x^2-10^2}</math> and <math>HG=\sqrt{y^2-10^2}</math> by the Pythagorean Theorem. | ||
+ | |||
+ | Since <math>\angle ABG</math> and <math>\angle HAG</math> are both complementary to <math>\angle AHB,</math> we have <math>\angle ABG = \angle HAG,</math> from which <math>\triangle ABG \sim \triangle HAG</math> by AA. It follows that <math>\frac{BG}{AG}=\frac{AG}{HG},</math> so <math>BG\cdot HG=AG^2,</math> or <cmath>\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)</cmath> | ||
+ | Since <math>\angle AHB = \angle FHD</math> by vertical angles, we have <math>\triangle AHB \sim \triangle FHD</math> by AA, with the ratio of similitude <math>\frac{AH}{FH}=\frac{BA}{DF}.</math> It follows that <math>DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.</math> | ||
+ | |||
+ | Since <math>\angle EBA = \angle ECD = \angle FDA</math> by angle chasing, we have <math>\triangle EBA \sim \triangle FDA</math> by AA, with the ratio of similitude <math>\frac{EA}{FA}=\frac{BA}{DA}.</math> It follows that <math>DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.</math> | ||
+ | |||
+ | By the Pythagorean Theorem on right <math>\triangle ADF,</math> we have <math>DF^2+AF^2=AD^2,</math> or <cmath>\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)</cmath> | ||
+ | Solving this system of equations (<math>(1)</math> and <math>(2)</math>), we get <math>x=\frac{45\sqrt2}{4}</math> and <math>y=\frac{90}{7},</math> so <math>AB=x=\frac{45\sqrt2}{4}</math> and <math>CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.</math> Finally, the area of <math>ABCD</math> is <cmath>K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},</cmath> from which <math>\sqrt2 \cdot K=\boxed{567}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=6rLnl8z7lnM | ||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=2021|n=I|num-b=8|num-a=10}} | {{AIME box|year=2021|n=I|num-b=8|num-a=10}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:20, 30 September 2021
Contents
Problem
Let be an isosceles trapezoid with and Suppose that the distances from to the lines and are and respectively. Let be the area of Find
Diagram
~MRENTHUSIASM
Solution 1
Construct your isosceles trapezoid. Let, for simplicity, , , and . Extend the sides and mark the intersection as . Following what the question states, drop a perpendicular from to labeling the foot as . Drop another perpendicular from to , calling the foot . Lastly, drop a perpendicular from to , labeling it . In addition, drop a perpendicular from to calling its foot .
Start out by constructing a triangle congruent to with its side of length on line . This works because all isosceles triangles are cyclic and as a result, .
Notice that by AA similarity. We are given that and by symmetry we can deduce that . As a result, . This gives us that .
The question asks us along the lines of finding the area, , of the trapezoid . We look at the area of and notice that it can be represented as . Substituting , we solve for , getting .
Now let us focus on isosceles triangle , where . Since, is an altitude from to of an isosceles triangle, must be equal to . Since and , we can solve to get that and .
We must then set up equations using the Pythagorean Theorem, writing everything in terms of , , and . Looking at right triangle we get Looking at right triangle we get Now rearranging and solving, we get two equation Those are convenient equations as which gives us After some "smart" calculation, we get that .
Notice that the question asks for , and by applying the trapezoid area formula. Fortunately, this is just , and plugging in the value of , we get that .
~Math_Genius_164
Solution 2 (LOC and Trig)
Let . Draw diagonal and let be the foot of the perpendicular from to , be the foot of the perpendicular from to line , and be the foot of the perpendicular from to .
Note that , and we get that . Therefore, . It then follows that . Using similar triangles, we can then find that . Using the Law of Cosines on , We can find that the . Since , and each is supplementary to , we know that the . It then follows that . Then it can be found that the area is . Multiplying this by , the answer is .
-happykeeper
Solution 3 (Similarity)
Let the foot of the altitude from to be , to be , and to be .
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, is on the circumcircle of and we have that is the Simson Line from . As , we have that , with the last equality coming from cyclic quadrilateral . Thus, and we have that or that , which we can see gives us that . Further ratios using the same similar triangles gives that and .
We also see that quadrilaterals and are both cyclic, with diameters of the circumcircles being and respectively. The intersection of the circumcircles are the points and , and we know and 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 taking to . Because we know a lot about but very little about 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 : we know that the altitude from to has length . As the two triangles are similar, if we can find the height from to , we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that . Using this, we can drop the altitude from to and let it intersect at . Then, let and thus . We then have by the Pythagorean Theorem on and : Then, . This gives us then from right triangle that and thus the ratio of to is . From this, we see then that and The Pythagorean Theorem on then gives that Then, we have the height of trapezoid is , the top base is , and the bottom base is . From the equation of a trapezoid, , so the answer is .
- lvmath
Solution 4 (Cool Solution by advanture)
First, draw the diagram. Then, notice that since is isosceles, , and the length of the altitude from to is also . Let the foot of this altitude be , and let the foot of the altitude from to be denoted as . Then, . So, . Now, notice that , where denotes the area of triangle . Letting , this equality becomes . Also, from , we have . Now, by the Pythagorean theorem on triangles and , we have and . Notice that , so . Squaring both sides of the equation once, moving and to the right, dividing both sides by , and squaring the equation once more, we are left with . Dividing both sides by (since we know is positive), we are left with . Solving for gives us .
Now, let the foot of the perpendicular from to be . Then let . Let the foot of the perpendicular from to be . Then, is also equal to . Notice that is a rectangle, so . Now, we have . By the Pythagorean theorem applied to , we have . We know that , so we can plug this into this equation. Solving for , we get .
Finally, to find , we use the formula for the area of a trapezoid: . The problem asks us for , which comes out to be .
~advanture
Solution 5 (Compact Similarity Solution)
Let and be the feet of the altitudes from to and , respectively.
Claim: We have pairs of similar right triangles: and .
Proof: Note that is cyclic. We need one more angle, and we get this from this cyclic quad:
Let . We obtain from the similarities and .
By Ptolemy, , so .
We obtain , so .
Applying the Pythagorean theorem on , we get .
Thus, , and , yielding .
Solution 6 (Two Variables, Two Equations)
Let and be the perpendiculars from to and respectively. Next, let be the intersection of and
We set and as shown below. From here, we obtain by segment subtraction, and and by the Pythagorean Theorem.
Since and are both complementary to we have from which by AA. It follows that so or Since by vertical angles, we have by AA, with the ratio of similitude It follows that
Since by angle chasing, we have by AA, with the ratio of similitude It follows that
By the Pythagorean Theorem on right we have or Solving this system of equations ( and ), we get and so and Finally, the area of is from which
~MRENTHUSIASM
Video Solution
https://www.youtube.com/watch?v=6rLnl8z7lnM
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
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