Octagon Problem

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High School Math Grades 9-12, Ages 13-18

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Octagon Problem

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22 posts

Shiyul

#1 h Apr 24, 2025, 11:41 PM

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The vertices of octagon $ABCDEFGH$ lie on the same circle. If $AB = BC = CD = DE = 11$ and $EF = FG = GH = HA = \sqrt2$ , what is the area of octagon $ABCDEFGH$ ?

I approached this problem by noticing that the area of the octagon is the area of the eight isoceles triangles with lengths $r$ , $r$ , and $sqrt2$ or 11. However, I didn't know how to find the radius. Can anyone give me a hint?

This post has been edited 1 time. Last edited by Shiyul, Apr 25, 2025, 6:47 PM
Reason: latex error

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Sid-darth-vater

43 posts

Sid-darth-vater

#2 h Apr 25, 2025, 12:08 AM

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Yeah, so what you want to do here is consider angles. They are extremely useful. Try to see what angles to consider yourself. If you can't find anything, look at these angles: Click to reveal hidden text

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22 posts

Shiyul

#3 h Apr 25, 2025, 12:39 AM

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Oh yeah, I forgot to say, but I tried that and I noticed that they add to 90, but I'm not sure what to do with that information.

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5851 posts

Sedro

#4 h Apr 25, 2025, 1:40 AM

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Building on @Sid-darth-vater's hint: idea

@below oh yeah, how did I miss that ...
you don't need $\angle AOH$ at all

This post has been edited 1 time. Last edited by Sedro, Apr 25, 2025, 1:52 AM

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Sid-darth-vater

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Sid-darth-vater

#5 h Apr 25, 2025, 1:43 AM • 1 Y

Y by Sedro

Shiyul wrote:

Oh yeah, I forgot to say, but I tried that and I noticed that they add to 90, but I'm not sure what to do with that information.

Adding to @Sedro look at the angles of the octagon itself, you might be able to determine an angle

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22 posts

Shiyul

#6 h Apr 25, 2025, 6:05 PM

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Okay. I solved the problem, an I'm pretty sure the answer is $121\sqrt3 + 8\sqrt{10}$ .

This post has been edited 1 time. Last edited by Shiyul, Apr 25, 2025, 6:06 PM
Reason: casdfasdfa

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Sid-darth-vater

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Sid-darth-vater

#7 h Apr 25, 2025, 6:59 PM

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hmm, I got an answer of Click to reveal hidden text, can I ask what you did?

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9058 posts

#8 h Apr 25, 2025, 7:35 PM

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The answer is $167$ , the radius is $\frac{\sqrt{290}}{2}$ .

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22 posts

Shiyul

#9 h Apr 25, 2025, 7:44 PM

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Can you tell me how you found that?

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9058 posts

#10 h Apr 25, 2025, 8:08 PM

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$\sin \alpha=\frac{\frac{11}{2}}{r}=\frac{11}{2r}$ and $\sin \beta=\frac{\frac{\sqrt{2}}{2}}{r}=\frac{\sqrt{2}}{2r}$ .

$4 \cdot (2\alpha+2\beta)=360^{\circ} \Rightarrow \alpha+\beta=45^{\circ}$ .
$\sin(\alpha+\beta)=\frac{\sqrt{2}}{2}$ ,
$\sin \alpha\cos \beta+\sin \beta\cos \alpha=\frac{\sqrt{2}}{2}$ ,
$\frac{11}{2r} \cdot \sqrt{1-(\frac{\sqrt{2}}{2})^{2}}+\frac{\sqrt{2}}{2r} \cdot \sqrt{1-(\frac{11}{2r})^{2}}=\frac{\sqrt{2}}{2} \quad (1)$ .
Squaring, calculating, again squaring, two solutions but after (1) only one solution.

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mathprodigy2011

342 posts

mathprodigy2011

#11 h Apr 25, 2025, 10:02 PM

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Shiyul wrote:

Oh yeah, I forgot to say, but I tried that and I noticed that they add to 90, but I'm not sure what to do with that information.

law of cosines could be useful, just my speculation, i havent rly solved the problem yet

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Sid-darth-vater

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Sid-darth-vater

#12 h Apr 26, 2025, 12:38 AM • 1 Y

Y by Sedro

Shiyul wrote:

Can you tell me how you found that?

Sure!fullsol also, I'm fairly new to writing proofs so please tell me if any improvements can be made, I'm trying to do better for next year!!

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