During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

2017 AMC 10B Problems/Problem 19

Problem

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3AB$. Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3BC$, and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3CA$. What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$? $\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1$

Solution

Solution 1

Note that by symmetry, $\triangle A'B'C'$ is also equilateral. Therefore, we only need to find one of the sides of $A'B'C'$ to determine the area ratio. WLOG, let $AB = BC = CA = 1$. Therefore, $BB' = 3$ and $BC' = 4$. Also, $\angle B'BC' = 120^{\circ}$, so by the Law of Cosines, $B'C' = \sqrt{37}$. Therefore, the answer is $(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37 : 1}$

Solution 2

As mentioned in the first solution, $\triangle A'B'C'$ is equilateral. WLOG, let $AB=2$. Let $D$ be on the line passing through $AB$ such that $A'D$ is perpendicular to $AB$. Note that $\triangle A'DA$ is a 30-60-90 with right angle at $D$. Since $AA'=6$, $AD=3$ and $A'D=3\sqrt{3}$. So we know that $DB'=11$. Note that $\triangle A'DB'$ is a right triangle with right angle at $D$. So by the Pythagorean theorem, we find $A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.$ Therefore, the answer is $(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37 : 1}$.

Solution 3

Let $AB=BC=CA=x$. We start by noting that we can just write $AB'$ as just $AB+BB'=4AB$. Similarly $BC'=4BC$, and $CA'=4CA$. We can evaluate the area of triangle $ABC$ by simply using Heron's formula, $[ABC]=\sqrt{\frac{3x}{2}\cdot {\Bigg(\frac{3x}{2}-x\Bigg)}^3}=\frac{x^2\sqrt{3}}{4}$. Next in order to evaluate $A'B'C'$ we need to evaluate the area of the larger triangles $AA'B',BB'C', \text{ and } CC'A'$. In this solution we shall just compute $1$ of these as the others are trivially equivalent. In order to compute the area of $\Delta{AA'B'}$ we can use the formula $[XYZ]=\frac{1}{2}xy\cdot\sin{z}$. Since $ABC$ is equilateral and $A$, $B$, $B'$ are collinear, we already know $\angle{A'AB'}=180-60=120$ Similarly from above we know $AB'$ and $A'A$ to be $4x$, and $3x$ respectively. Thus the area of $\Delta{AA'B'}$ is $\frac{1}{2}\cdot 4x\cdot 3x \cdot \sin{120}=3x^2\cdot\sqrt{3}$. Likewise we can find $BB'C', \text{ and } CC'A'$ to also be $3x^2\cdot\sqrt{3}$. $[A'B'C']=[AA'B']+[BB'C']+[CC'A']+[ABC]=3\cdot3x^2\cdot\sqrt{3}+\frac{x\sqrt{3}}{2}=\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)$. Therefore the ratio of $[A'B'C']$ to $[ABC]$ is $\frac{\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37 : 1}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
{"ft_lang_label":"__label__en","ft_lang_prob":0.82640105,"math_prob":1.0000054,"size":2048,"snap":"2022-05-2022-21","text_gpt3_token_len":593,"char_repetition_ratio":0.12328767,"word_repetition_ratio":0.2725225,"special_character_ratio":0.33740234,"punctuation_ratio":0.120192304,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000095,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,4,null,4,null,4,null,4,null,null,null,4,null,null,null,null,null,null,null,null,null,4,null,null,null,4,null,null,null,4,null,null,null,4,null,4,null,4,null,null,null,4,null,4,null,null,null,null,null,2,null,2,null,2,null,null,null,2,null,null,null,2,null,null,null,4,null,2,null,null,null,null,null,null,null,null,null,2,null,null,null,2,null,null,null,null,null,4,null,2,null,2,null,2,null,2,null,2,null,null,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-22T09:47:23Z\",\"WARC-Record-ID\":\"<urn:uuid:0cdccace-049f-455c-afbb-88fc1d3dfc91>\",\"Content-Length\":\"57031\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:818beea3-bc87-44fc-a9d1-0344347eef32>\",\"WARC-Concurrent-To\":\"<urn:uuid:8acca377-f70a-4618-8d61-cb85644804f7>\",\"WARC-IP-Address\":\"104.26.11.229\",\"WARC-Target-URI\":\"https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_19&direction=next&oldid=84049\",\"WARC-Payload-Digest\":\"sha1:AD5DRRF6JTNS4ABOFEIHTGG65PMA5BOO\",\"WARC-Block-Digest\":\"sha1:BNUU6QOKYBMUOGIKHJKPSG47JUZB63YH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303779.65_warc_CC-MAIN-20220122073422-20220122103422-00073.warc.gz\"}"}