Mixing
The set of all possible values of $AC$ is an open interval $(m,n)$
$begingroup$
Side $overline{AB}$ of $triangle ABC$ has length $10$. The bisector of angle $A$ meets $overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?
A) 16 B) 17 C) 18 D) 19 E) 20
Could someone help me understand what is wrong with my approach (none of the answers matches mine!):
Letting BD equal $x$ (which must be positive) and using the angle bisector theorem, we get that AC is $frac{30}{x}$. Using triangular inequality, we get that $$7 < x+frac{30}{x}$$ $$x<7+frac{30}{x}$$ and $$frac{30}{x} < 13+x.$$
We can multiply $x$ to both sides of all 3 inequalities because $x>0$ and will not change the direction of the inequality. Doing so and moving all terms to the LHS, we get $x^2-77x+30>0$ (1), $x^2-7x-30<0$ (2) and $x^2+13x-30>0$ (3). (1) is always true because we can re-write it in the form of $(x-a)^2+b>0$ where $a$ and $b$ are both positive. (2) can be rewritten as $(x+3)(x-10)<0$, and thus $-3<x<10$. But $x>0$, so $0<x<10$.
Finally, (3) can also be written as $(x+15)(x-2)>0$ where only $x>2$ works because $x<-15$ is less than 0. Then combining all the scenarios, $x$ can range from $2$ to $10$, but $12$ is not an option... I've double/triple checked my work, but there must be something I'm missing.
contest-math euclidean-geometry interval-arithmetic
edited Jan 30 at 7:44
YuiTo Cheng
2,1863937
asked Jan 30 at 6:33
jjhhjjhh
2,14411123
$endgroup$
add a comment |
$begingroup$
Side $overline{AB}$ of $triangle ABC$ has length $10$. The bisector of angle $A$ meets $overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?
A) 16 B) 17 C) 18 D) 19 E) 20
Could someone help me understand what is wrong with my approach (none of the answers matches mine!):
Letting BD equal $x$ (which must be positive) and using the angle bisector theorem, we get that AC is $frac{30}{x}$. Using triangular inequality, we get that $$7 < x+frac{30}{x}$$ $$x<7+frac{30}{x}$$ and $$frac{30}{x} < 13+x.$$
We can multiply $x$ to both sides of all 3 inequalities because $x>0$ and will not change the direction of the inequality. Doing so and moving all terms to the LHS, we get $x^2-77x+30>0$ (1), $x^2-7x-30<0$ (2) and $x^2+13x-30>0$ (3). (1) is always true because we can re-write it in the form of $(x-a)^2+b>0$ where $a$ and $b$ are both positive. (2) can be rewritten as $(x+3)(x-10)<0$, and thus $-3<x<10$. But $x>0$, so $0<x<10$.
Finally, (3) can also be written as $(x+15)(x-2)>0$ where only $x>2$ works because $x<-15$ is less than 0. Then combining all the scenarios, $x$ can range from $2$ to $10$, but $12$ is not an option... I've double/triple checked my work, but there must be something I'm missing.
contest-math euclidean-geometry interval-arithmetic
edited Jan 30 at 7:44
YuiTo Cheng
2,1863937
asked Jan 30 at 6:33
jjhhjjhh
2,14411123
$endgroup$
$begingroup$
This is a minor point, but there is a typo in your $left(1right)$ equation, where $- 77x$ should be $-7x$.
$endgroup$
– John Omielan
Jan 30 at 6:59
$begingroup$
@JohnOmielan — ah, right. Thank you for pointing that out!
$endgroup$
– jjhh
Jan 31 at 1:18
add a comment |
$begingroup$
Side $overline{AB}$ of $triangle ABC$ has length $10$. The bisector of angle $A$ meets $overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?
A) 16 B) 17 C) 18 D) 19 E) 20
Could someone help me understand what is wrong with my approach (none of the answers matches mine!):
Letting BD equal $x$ (which must be positive) and using the angle bisector theorem, we get that AC is $frac{30}{x}$. Using triangular inequality, we get that $$7 < x+frac{30}{x}$$ $$x<7+frac{30}{x}$$ and $$frac{30}{x} < 13+x.$$
We can multiply $x$ to both sides of all 3 inequalities because $x>0$ and will not change the direction of the inequality. Doing so and moving all terms to the LHS, we get $x^2-77x+30>0$ (1), $x^2-7x-30<0$ (2) and $x^2+13x-30>0$ (3). (1) is always true because we can re-write it in the form of $(x-a)^2+b>0$ where $a$ and $b$ are both positive. (2) can be rewritten as $(x+3)(x-10)<0$, and thus $-3<x<10$. But $x>0$, so $0<x<10$.
Finally, (3) can also be written as $(x+15)(x-2)>0$ where only $x>2$ works because $x<-15$ is less than 0. Then combining all the scenarios, $x$ can range from $2$ to $10$, but $12$ is not an option... I've double/triple checked my work, but there must be something I'm missing.
contest-math euclidean-geometry interval-arithmetic
edited Jan 30 at 7:44
YuiTo Cheng
2,1863937
asked Jan 30 at 6:33
jjhhjjhh
2,14411123
$endgroup$
Side $overline{AB}$ of $triangle ABC$ has length $10$. The bisector of angle $A$ meets $overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?
A) 16 B) 17 C) 18 D) 19 E) 20
Could someone help me understand what is wrong with my approach (none of the answers matches mine!):
Letting BD equal $x$ (which must be positive) and using the angle bisector theorem, we get that AC is $frac{30}{x}$. Using triangular inequality, we get that $$7 < x+frac{30}{x}$$ $$x<7+frac{30}{x}$$ and $$frac{30}{x} < 13+x.$$
We can multiply $x$ to both sides of all 3 inequalities because $x>0$ and will not change the direction of the inequality. Doing so and moving all terms to the LHS, we get $x^2-77x+30>0$ (1), $x^2-7x-30<0$ (2) and $x^2+13x-30>0$ (3). (1) is always true because we can re-write it in the form of $(x-a)^2+b>0$ where $a$ and $b$ are both positive. (2) can be rewritten as $(x+3)(x-10)<0$, and thus $-3<x<10$. But $x>0$, so $0<x<10$.
Finally, (3) can also be written as $(x+15)(x-2)>0$ where only $x>2$ works because $x<-15$ is less than 0. Then combining all the scenarios, $x$ can range from $2$ to $10$, but $12$ is not an option... I've double/triple checked my work, but there must be something I'm missing.
contest-math euclidean-geometry interval-arithmetic
contest-math euclidean-geometry interval-arithmetic
edited Jan 30 at 7:44
YuiTo Cheng
2,1863937
asked Jan 30 at 6:33
jjhhjjhh
2,14411123
edited Jan 30 at 7:44
YuiTo Cheng
2,1863937
asked Jan 30 at 6:33
jjhhjjhh
2,14411123
edited Jan 30 at 7:44
YuiTo Cheng
2,1863937
edited Jan 30 at 7:44
YuiTo Cheng
2,1863937
edited Jan 30 at 7:44
YuiTo Cheng
2,1863937
2,1863937
asked Jan 30 at 6:33
jjhhjjhh
2,14411123
asked Jan 30 at 6:33
jjhhjjhh
2,14411123
asked Jan 30 at 6:33
jjhhjjhh
2,14411123
2,14411123
$begingroup$
This is a minor point, but there is a typo in your $left(1right)$ equation, where $- 77x$ should be $-7x$.
$endgroup$
– John Omielan
Jan 30 at 6:59
$begingroup$
@JohnOmielan — ah, right. Thank you for pointing that out!
$endgroup$
– jjhh
Jan 31 at 1:18
add a comment |
$begingroup$
This is a minor point, but there is a typo in your $left(1right)$ equation, where $- 77x$ should be $-7x$.
$endgroup$
– John Omielan
Jan 30 at 6:59
$begingroup$
@JohnOmielan — ah, right. Thank you for pointing that out!
$endgroup$
– jjhh
Jan 31 at 1:18
$begingroup$
This is a minor point, but there is a typo in your $left(1right)$ equation, where $- 77x$ should be $-7x$.
$endgroup$
– John Omielan
Jan 30 at 6:59
$begingroup$
This is a minor point, but there is a typo in your $left(1right)$ equation, where $- 77x$ should be $-7x$.
$endgroup$
– John Omielan
Jan 30 at 6:59
$begingroup$
@JohnOmielan — ah, right. Thank you for pointing that out!
$endgroup$
– jjhh
Jan 31 at 1:18
$begingroup$
@JohnOmielan — ah, right. Thank you for pointing that out!
$endgroup$
– jjhh
Jan 31 at 1:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I believe you have the correct range for $x$ of $2 lt x lt 10$, but that is for $overline{BD}$. However, the question asks for $overline{AC}$. Since $overline{AC} = frac{30}{x}$, this means it's range is $3 lt overline{AC} lt 15$, so $m = 3$ and $n = 15$, giving the correct answer of $m + n = 3 + 15 = 18$, i.e., choice $C$.
answered Jan 30 at 6:53
John OmielanJohn Omielan
4,6012215
$endgroup$
$begingroup$
Okay, I see this was a pretty careless mistake now... Perhaps I should have set AC into $x$ instead.
$endgroup$
– jjhh
Jan 31 at 1:19
$begingroup$
@jjhh Perhaps it would have been better to have $overline{AC} = x$ instead, with $overline{BD} = frac{30}{x}$ in that case. IMO, it's usually not too important which values you set to be specific variables, as long as you keep track of what you're trying to determine. However, although it doesn't make too much difference here, in some more complicated problems, using the appropriate variables can help make the calculations much easier & simpler than they could be otherwise.
$endgroup$
– John Omielan
Jan 31 at 1:25
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3093173%2fthe-set-of-all-possible-values-of-ac-is-an-open-interval-m-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe you have the correct range for $x$ of $2 lt x lt 10$, but that is for $overline{BD}$. However, the question asks for $overline{AC}$. Since $overline{AC} = frac{30}{x}$, this means it's range is $3 lt overline{AC} lt 15$, so $m = 3$ and $n = 15$, giving the correct answer of $m + n = 3 + 15 = 18$, i.e., choice $C$.
answered Jan 30 at 6:53
John OmielanJohn Omielan
4,6012215
$endgroup$
$begingroup$
Okay, I see this was a pretty careless mistake now... Perhaps I should have set AC into $x$ instead.
$endgroup$
– jjhh
Jan 31 at 1:19
$begingroup$
@jjhh Perhaps it would have been better to have $overline{AC} = x$ instead, with $overline{BD} = frac{30}{x}$ in that case. IMO, it's usually not too important which values you set to be specific variables, as long as you keep track of what you're trying to determine. However, although it doesn't make too much difference here, in some more complicated problems, using the appropriate variables can help make the calculations much easier & simpler than they could be otherwise.
$endgroup$
– John Omielan
Jan 31 at 1:25
add a comment |
$begingroup$
I believe you have the correct range for $x$ of $2 lt x lt 10$, but that is for $overline{BD}$. However, the question asks for $overline{AC}$. Since $overline{AC} = frac{30}{x}$, this means it's range is $3 lt overline{AC} lt 15$, so $m = 3$ and $n = 15$, giving the correct answer of $m + n = 3 + 15 = 18$, i.e., choice $C$.
answered Jan 30 at 6:53
John OmielanJohn Omielan
4,6012215
$endgroup$
$begingroup$
Okay, I see this was a pretty careless mistake now... Perhaps I should have set AC into $x$ instead.
$endgroup$
– jjhh
Jan 31 at 1:19
$begingroup$
@jjhh Perhaps it would have been better to have $overline{AC} = x$ instead, with $overline{BD} = frac{30}{x}$ in that case. IMO, it's usually not too important which values you set to be specific variables, as long as you keep track of what you're trying to determine. However, although it doesn't make too much difference here, in some more complicated problems, using the appropriate variables can help make the calculations much easier & simpler than they could be otherwise.
$endgroup$
– John Omielan
Jan 31 at 1:25
add a comment |
$begingroup$
I believe you have the correct range for $x$ of $2 lt x lt 10$, but that is for $overline{BD}$. However, the question asks for $overline{AC}$. Since $overline{AC} = frac{30}{x}$, this means it's range is $3 lt overline{AC} lt 15$, so $m = 3$ and $n = 15$, giving the correct answer of $m + n = 3 + 15 = 18$, i.e., choice $C$.
answered Jan 30 at 6:53
John OmielanJohn Omielan
4,6012215
$endgroup$
I believe you have the correct range for $x$ of $2 lt x lt 10$, but that is for $overline{BD}$. However, the question asks for $overline{AC}$. Since $overline{AC} = frac{30}{x}$, this means it's range is $3 lt overline{AC} lt 15$, so $m = 3$ and $n = 15$, giving the correct answer of $m + n = 3 + 15 = 18$, i.e., choice $C$.
answered Jan 30 at 6:53
John OmielanJohn Omielan
4,6012215
edited Jan 30 at 7:04
answered Jan 30 at 6:53
John OmielanJohn Omielan
4,6012215
answered Jan 30 at 6:53
John OmielanJohn Omielan
4,6012215
answered Jan 30 at 6:53
John OmielanJohn Omielan
4,6012215
4,6012215
$begingroup$
Okay, I see this was a pretty careless mistake now... Perhaps I should have set AC into $x$ instead.
$endgroup$
– jjhh
Jan 31 at 1:19
$begingroup$
@jjhh Perhaps it would have been better to have $overline{AC} = x$ instead, with $overline{BD} = frac{30}{x}$ in that case. IMO, it's usually not too important which values you set to be specific variables, as long as you keep track of what you're trying to determine. However, although it doesn't make too much difference here, in some more complicated problems, using the appropriate variables can help make the calculations much easier & simpler than they could be otherwise.
$endgroup$
– John Omielan
Jan 31 at 1:25
add a comment |
$begingroup$
Okay, I see this was a pretty careless mistake now... Perhaps I should have set AC into $x$ instead.
$endgroup$
– jjhh
Jan 31 at 1:19
$begingroup$
@jjhh Perhaps it would have been better to have $overline{AC} = x$ instead, with $overline{BD} = frac{30}{x}$ in that case. IMO, it's usually not too important which values you set to be specific variables, as long as you keep track of what you're trying to determine. However, although it doesn't make too much difference here, in some more complicated problems, using the appropriate variables can help make the calculations much easier & simpler than they could be otherwise.
$endgroup$
– John Omielan
Jan 31 at 1:25
$begingroup$
Okay, I see this was a pretty careless mistake now... Perhaps I should have set AC into $x$ instead.
$endgroup$
– jjhh
Jan 31 at 1:19
$begingroup$
Okay, I see this was a pretty careless mistake now... Perhaps I should have set AC into $x$ instead.
$endgroup$
– jjhh
Jan 31 at 1:19
$begingroup$
@jjhh Perhaps it would have been better to have $overline{AC} = x$ instead, with $overline{BD} = frac{30}{x}$ in that case. IMO, it's usually not too important which values you set to be specific variables, as long as you keep track of what you're trying to determine. However, although it doesn't make too much difference here, in some more complicated problems, using the appropriate variables can help make the calculations much easier & simpler than they could be otherwise.
$endgroup$
– John Omielan
Jan 31 at 1:25
$begingroup$
@jjhh Perhaps it would have been better to have $overline{AC} = x$ instead, with $overline{BD} = frac{30}{x}$ in that case. IMO, it's usually not too important which values you set to be specific variables, as long as you keep track of what you're trying to determine. However, although it doesn't make too much difference here, in some more complicated problems, using the appropriate variables can help make the calculations much easier & simpler than they could be otherwise.
$endgroup$
– John Omielan
Jan 31 at 1:25
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3093173%2fthe-set-of-all-possible-values-of-ac-is-an-open-interval-m-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
This is a minor point, but there is a typo in your $left(1right)$ equation, where $- 77x$ should be $-7x$.
$endgroup$
– John Omielan
Jan 30 at 6:59
$begingroup$
@JohnOmielan — ah, right. Thank you for pointing that out!
$endgroup$
– jjhh
Jan 31 at 1:18