Difference between revisions of "1992 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
+ | In triangle <math>ABC^{}_{}</math>, <math>A'</math>, <math>B'</math>, and <math>C'</math> are on the sides <math>BC</math>, <math>AC^{}_{}</math>, and <math>AB^{}_{}</math>, respectively. Given that <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92</math>, find <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}</math>. | ||
− | == Solution == | + | == Solution 1== |
− | {{ | + | Let <math>K_A=[BOC], K_B=[COA],</math> and <math>K_C=[AOB].</math> Due to triangles <math>BOC</math> and <math>ABC</math> having the same base, <cmath>\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.</cmath> Therefore, we have |
+ | <cmath>\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}</cmath> <cmath>\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}</cmath> <cmath>\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}.</cmath> | ||
+ | Thus, we are given <cmath>\frac{K_B+K_C}{K_A}+\frac{K_A+K_C}{K_B}+\frac{K_A+K_B}{K_C}=92.</cmath> Combining and expanding gives <cmath>\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}=92.</cmath> We desire <math>\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_AK_BK_C}.</math> Expanding this gives <cmath>\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}+2=\boxed{094}.</cmath> | ||
+ | |||
+ | == Solution 2 == | ||
+ | Using [[mass points]], let the weights of <math>A</math>, <math>B</math>, and <math>C</math> be <math>a</math>, <math>b</math>, and <math>c</math> respectively. | ||
+ | |||
+ | Then, the weights of <math>A'</math>, <math>B'</math>, and <math>C'</math> are <math>b+c</math>, <math>c+a</math>, and <math>a+b</math> respectively. | ||
+ | |||
+ | Thus, <math>\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}</math>, <math>\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}</math>, and <math>\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}</math>. | ||
+ | |||
+ | Therefore: | ||
+ | <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'} = \frac{b+c}{a} \cdot \frac{c+a}{b} \cdot \frac{a+b}{c}</math> <math>= \frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =</math> | ||
+ | |||
+ | <math>2+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}+\frac{ab(a+b)}{abc} = 2 + \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} </math> <math>= 2 + \frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'} = 2+92 = \boxed{094}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | As in above solutions, find <math>\sum_{cyc} \frac{y+z}{x}=92</math> (where <math>O=(x:y:z)</math> in barycentric coordinates). Now letting <math>y=z=1</math> we get <math>\frac{2}{x}+2(x+1)=92 \implies x+\frac{1}{x}=45</math>, and so <math>\frac{2}{x}(x+1)^2=2(x+\frac{1}{x}+2)=2 \cdot 47 = 94</math>. | ||
+ | |||
+ | ~Lcz | ||
+ | |||
+ | == Solution 4 (Ceva's Theorem) == | ||
+ | |||
+ | A consequence of Ceva's theorem sometimes attributed to Gergonne is that <math>\frac{AO}{OA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}</math>, and similarly for cevians <math>BB'</math> and <math>CC'</math>. Now we apply Gergonne several times and do algebra: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \frac{AO}{OA'}\frac{BO}{OB'}\frac{CO}{OC'} &= | ||
+ | \left(\frac{AB'}{B'C}+\frac{AC'}{C'B}\right) | ||
+ | \left(\frac{BC'}{C'A}+\frac{BA'}{A'C}\right) | ||
+ | \left(\frac{CB'}{B'A}+\frac{CA'}{A'B}\right)\ | ||
+ | &=\underbrace{\frac{AB'\cdot CA'\cdot BC'}{B'C\cdot A'B\cdot C'A}}_{\text{Ceva}} + | ||
+ | \underbrace{\frac{AC'\cdot BA'\cdot CB'}{C'B\cdot A'C\cdot B'A}}_{\text{Ceva}} + | ||
+ | \underbrace{\frac{AB'}{B'C} + \frac{AC'}{C'B}}_{\text{Gergonne}} + | ||
+ | \underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{\text{Gergonne}} + | ||
+ | \underbrace{\frac{CA'}{A'B} + \frac{CB'}{B'A}}_{\text{Gergonne}}\ | ||
+ | &= 1 + 1 + \underbrace{\frac{AO}{OA'} + \frac{BO}{OB'} + \frac{CO}{OC'}}_{92} = \boxed{94} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~ proloto | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1992|num-b=13|num-a=15}} | |
− | |||
− | |||
− | + | [[Category:Intermediate Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 00:26, 16 August 2023
Contents
[hide]Problem
In triangle , , , and are on the sides , , and , respectively. Given that , , and are concurrent at the point , and that , find .
Solution 1
Let and Due to triangles and having the same base, Therefore, we have Thus, we are given Combining and expanding gives We desire Expanding this gives
Solution 2
Using mass points, let the weights of , , and be , , and respectively.
Then, the weights of , , and are , , and respectively.
Thus, , , and .
Therefore:
.
Solution 3
As in above solutions, find (where in barycentric coordinates). Now letting we get , and so .
~Lcz
Solution 4 (Ceva's Theorem)
A consequence of Ceva's theorem sometimes attributed to Gergonne is that , and similarly for cevians and . Now we apply Gergonne several times and do algebra:
~ proloto
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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