# Calculate the largest integer value of segment $CE$ . in the triangle below.

Calculate the largest integral value of $$CE$$ where $$AC$$ has the smallest integral value and $$BC = 3m$$ and $$AC = CD$$ ($$\alpha>90^\circ$$)

(Answer:$$8$$)

$$\angle ACD = 180^\circ - 4\theta \implies \angle CAD \cong \angle CDA = 2\theta$$

$$\therefore \angle CED = \theta \implies \triangle CDE_{\text{(isosc)}}$$

$$\therefore CD = ED$$

$$\triangle ABC: AC^2 = AB^2+3^2-2.3.AB.\cos\alpha = AB^2+9+6\cos\alpha(90^\circ < \alpha < 180^\circ )$$

$$\triangle CDE: CE^2 = CD^2+DE^2-2.CD.DE.\cos(180-2\theta) = 2AC^2-2AC^2\cos(180-2\theta)$$

$$\therefore CE^2 = AC^2 (2-\cos(180^\circ -2\theta)) = AC^2(2-(-\cos(2\theta))$$

$$\therefore CE^2 = AC^2(2+\cos2\theta)$$

How do I finish?

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Commented Aug 2 at 16:44

First, for simpler algebra, let

$$j = \lvert AC\rvert, \;\; k = \lvert CE\rvert, \;\; j, k \in \mathbb{N}$$

Since $$\lvert AC\rvert = \lvert CD\rvert = \lvert DE\rvert = j$$, with $$\measuredangle DCE = \theta$$, as you've determined, then $$\triangle CDE$$ is isosceles. Thus, the perpendicular bisector of $$CE$$ is also the angle bisector of $$\angle CDE$$, so half of $$\lvert CE\rvert$$ is $$\lvert CD\rvert\cos\theta$$, i.e.,

$$\lvert CE\rvert = 2\lvert CD\rvert\cos\theta \;\;\to\;\; k = 2j\cos\theta$$

If $$\theta = 0$$, then $$\cos\theta = 1$$ so $$k = 2j$$. As $$j \gt 3$$ (due to $$\alpha \gt 90^{\circ}$$, the smallest possible value of $$j$$ is $$4$$), we get $$j = 4$$ and $$k = 8$$, which matches the current question stated solution. However, $$\triangle ABE$$ is then a degenerate triangle, where it's just the one line segment of $$AE$$.

Otherwise, with $$\theta \gt 0$$, then $$\cos\theta \lt 1$$ so $$k \lt 2j$$. Thus, for $$k$$ to be an integer means that $$k \le 2j - 1$$. To determine the smallest values of $$j$$ and $$k$$, consider the limiting minimum case where $$\alpha = \alpha_m = 90^{\circ}$$. The law of cosines in $$\triangle ABC$$ gives

$$\lvert AC\rvert^2 = \lvert AB\rvert^2 + \lvert BC\rvert^2 - 2\lvert AB\rvert\lvert BC\rvert\cos(\measuredangle ABC) \;\to\; j^2 = \lvert AB\rvert^2 + 9 - 6\lvert AB\rvert\cos(\alpha)$$

With $$\alpha$$ increasing more than $$90^{\circ}$$, note $$\cos(\alpha)$$ goes from $$0$$ to increasingly negative values. If $$\lvert AB\rvert$$ were to stay the same or increase, the RHS (and thus also the LHS) value would increase. However, the LHS is constant, so $$\lvert AB\rvert$$ must decrease. Using the law of cosines again, we have

$$\lvert AB\rvert^2 = \lvert AC\rvert^2 + \lvert BC\rvert^2 - 2\lvert AC\rvert\lvert BC\rvert\cos(\measuredangle BCA) = j^2 + 9 - 6j\cos(3\theta)$$

As $$\lvert AB\rvert$$ is decreasing, this means since $$3\theta \lt 90^{\circ}$$ that $$\cos(3\theta) \gt 0$$, so $$\cos(3\theta)$$ must be increasing, i.e., $$3\theta$$ is decreasing. Thus, if $$\theta_u$$ is the value of $$\theta$$ with $$\alpha = 90^{\circ}$$, then $$\theta_u$$ is the upper bound of the possible values of $$\theta$$, i.e.,

$$\theta \lt \theta_u \;\;\to\;\; \cos\theta \gt \cos\theta_u$$

With $$\alpha = 90^{\circ}$$, then $$\triangle ABC$$ is right-angled, so we get

$$\cos(3\theta_u) = \frac{\lvert BC\rvert}{\lvert AC\rvert} = \frac{3}{j} \;\;\to\;\; \theta_u = \frac{\cos^{-1}\left(\frac{3}{j}\right)}{3}$$

With $$k_l$$ being the corresponding value of $$\lvert CD\rvert$$, since $$\cos(\theta)$$ is increasing, we also have

$$2j - 1 \ge k \gt k_l = 2j\cos\theta_u$$

where $$k_l$$ is lower bound of the possible values of $$k$$. The following table shows the values of $$j$$ starting from $$4$$, along with approximations of $$\theta_u$$ and $$k_l$$, to show which one is the first to have a potentially valid value of $$k$$, i.e., where $$k_l \le 2j - 1$$.

$$j$$ $$\theta_u \approx$$ $$k_l \approx$$ $$2j - 1$$ $$k$$
$$4$$ $$13.803^{\circ}$$ $$7.769$$ $$7$$ N.A.
$$5$$ $$17.710^{\circ}$$ $$9.526$$ $$9$$ N.A.
$$6$$ $$20^{\circ}$$ $$11.276$$ $$11$$ N.A.
$$7$$ $$21.541^{\circ}$$ $$13.022$$ $$13$$ N.A.
$$8$$ $$22.659^{\circ}$$ $$14.765$$ $$15$$ $$15$$

This shows the smallest potential value of $$\lvert CE\rvert$$ is $$15$$, corresponding to $$\lvert AC\rvert = 8$$. To confirm this is the case, i.e., that everything works properly with this value, we have

$$\theta = \cos^{-1}\left(\frac{15}{16}\right) \approx 20.364^{\circ} \;\;\to\;\; 3\theta \approx 61.092^{\circ}$$

Using the law of cosines with $$\triangle ABC$$, we get

\begin{aligned} \lvert AB\rvert^2 & = 3^2 + 8^2 - 2(3)(8)\cos(3\theta) \\ \lvert AB\rvert^2 & = 73 - 48\cos(3\theta) \\ \lvert AB\rvert^2 & \approx 49.797 \\ \lvert AB\rvert & \approx 7.057 \end{aligned}

Next, the law of sines applied to $$\triangle ABC$$ results in

\begin{aligned} \frac{\sin(\measuredangle BAC)}{\lvert BC\rvert} & = \frac{\sin(\measuredangle BCA)}{\lvert AB\rvert} \\ \frac{\sin(\measuredangle BAC)}{3} & \approx \frac{\sin(61.092^{\circ})}{7.057} \\ \sin(\measuredangle BAC) & \approx 0.372 \\ \measuredangle BAC & \approx 21.849^{\circ} \end{aligned}

Finally, there's

$$\alpha = 180^{\circ} - 3\theta - \measuredangle BAC \approx 180^{\circ} - 61.092^{\circ} - 21.849^{\circ} = 97.059^{\circ} \gt 90^{\circ}$$

Thus, everything is consistent.

Note the current question stated solution of $$8$$ matches only the degenerate case. However, the original stated solution of $$7$$ doesn't match what I got above, although $$15 - 8 = 7$$, i.e., the difference in values is $$7$$. Apart from this, there's only one other fairly only reasonable possibility I know of such that $$7$$ would be the correct answer. This is where, although it's not very common, and I consider it to be somewhat poor practice, I've seen people say on several occasions that the "integer value" of something like $$3.45$$ is $$3$$, i.e., it means the "integer part" or floor of the value instead. Although you've stated this is not the intended meaning, to show that $$7$$ does result in this case, note that since $$\lvert AC\rvert \gt 3$$, the smallest possible integer part is $$3$$. Also, since $$\lvert CE\rvert \lt 8$$, the latest possible integer part of it is $$7$$.

To get a specific example, have $$\theta = 1^{\circ} \;\to\; 3\theta = 3^{\circ}$$ and $$\lvert CE\rvert = 7.01$$ to get

$$\lvert AC\rvert = \frac{\lvert CE\rvert}{2\cos\theta} = \frac{7.01}{2\cos(1^{\circ})} \approx 3.5055$$

Applying the same procedure as used above for $$j = 8$$ and $$k = 15$$, we get using the law of cosines with $$\triangle ABC$$ that

\begin{aligned} \lvert AB\rvert^2 & \approx 3^2 + 3.5055^2 - 2(3)(3.5055)\cos(3\theta) \\ \lvert AB\rvert^2 & \approx 21.2888 - 21.0332\cos(3\theta) \\ \lvert AB\rvert^2 & \approx 0.2844 \\ \lvert AB\rvert & \approx 0.5333 \end{aligned}

The law of sines with $$\triangle ABC$$ then has

\begin{aligned} \frac{\sin(\measuredangle BAC)}{\lvert BC\rvert} & = \frac{\sin(\measuredangle BCA)}{\lvert AB\rvert} \\ \frac{\sin(\measuredangle BAC)}{3} & \approx \frac{\sin(3^{\circ})}{0.5333} \\ \sin(\measuredangle BAC) & \approx 0.2944 \\ \measuredangle BAC & \approx 17.123^{\circ} \end{aligned}

Finally, we have

$$\alpha = 180^{\circ} - 3\theta - \measuredangle BAC \approx 180^{\circ} - 3^{\circ} - 17.123^{\circ} = 159.877^{\circ} \gt 90^{\circ}$$

Thus, everything is consistent with the expected answer of $$7$$.

• I made the change as suggested. It's funny how the interpretations are different. For my country, it's very clear what is being asked. For the smallest integer value that angle AC can assume, find the largest integer value that CE can assume. Ex: If AC > 9.2, the smallest integer value would be 10, and if the value of DE for this value of AC were DE < 12.5, then the largest integer value would be 12. We don't use the integral term here. And for the floor and ceiling, we always use the floor function and ceiling function. Integer values ​​are the integers ($n \in Z$). Commented Aug 2 at 0:26
• continue...I didn't quite understand your solution. I found it complicated... I imagine the question is for a simpler solution like the one another colleague posted in the chat. Cheers Commented Aug 2 at 0:26
• @petaarantes Note that, most of the time, "integer value" does mean exactly what you state in the situations I'm aware of. It's only on several occasions, but not very often and I don't think it's particularly appropriate, I've seen "integer value" being used the same as "integer part". The only reason I mentioned it is because the stated solution of $7$ doesn't work with that interpretation, but it does with the alternate interpretation, so I thought perhaps that was intended. Regarding my solution being "complicated", that's partially because we need to ensure that everything is ... Commented Aug 2 at 0:32
• @petaarantes (cont.) consistent with whatever the proposed solution is. If there's anything in particular that you would like me to explain in more detail, please let me know. Finally, as for "... the question is for a simpler solution like the one another colleague posted in the chat", it's possible there's an overall simpler solution. However, if the chat solution is the the one you referred to here, i.e., $\lvert AC\rvert=4$ & $\lvert CE\rvert=7$, ... Commented Aug 2 at 0:36
• @petaarantes (cont.) it's not a valid solution, as I explained in my next comment, and have also shown in my answer above. If you're referring to a different chat solution, one that's valid, I'd like to know about it, especially if it's simpler and/or shorter that what I've done above. Commented Aug 2 at 0:38